Datrys x+2/4-[2(1-x)]^2/64=-(x+1/4)^2+(1-1/2)x+frac{1}{2}

$$\frac { x + 2 } { 4 } - \frac { [ 2 ( 1 - x ) ] ^ { 2 } } { 64 } = - ( \frac { x + 1 } { 4 } ) ^ { 2 } + ( 1 - \frac { 1 } { 2 } ) x + \frac { 1 } { 2 }$$

Datrys ar gyfer x (complex solution)

$$x\in \mathrm{C}$$

Camau Ateb Byr

Lluoswch ddwy ochr yr hafaliad wrth 64, lluoswm cyffredin lleiaf 4,64,2.

$$16\left(x+2\right)-\left(2\left(1-x\right)\right)^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

Defnyddio’r briodwedd ddosbarthu i luosi 16 â x+2.

$$16x+32-\left(2\left(1-x\right)\right)^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

Defnyddio’r briodwedd ddosbarthu i luosi 2 â 1-x.

$$16x+32-\left(2-2x\right)^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

Defnyddio'r theorem binomaidd \left(a-b\right)^{2}=a^{2}-2ab+b^{2} i ehangu'r \left(2-2x\right)^{2}.

$$16x+32-\left(4-8x+4x^{2}\right)=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

I ddod o hyd i wrthwyneb 4-8x+4x^{2}, dewch o hyd i wrthwyneb pob term.

$$16x+32-4+8x-4x^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

Tynnu 4 o 32 i gael 28.

$$16x+28+8x-4x^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

Cyfuno 16x a 8x i gael 24x.

$$24x+28-4x^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

I godi \frac{x+1}{4} i bŵer, codwch y rhifiadur a'r enwadur i bŵer ac yna rhannwch nhw.

$$24x+28-4x^{2}=-64\times \frac{\left(x+1\right)^{2}}{4^{2}}+64\left(1-\frac{1}{2}\right)x+32$$

Mynegwch -64\times \frac{\left(x+1\right)^{2}}{4^{2}} fel ffracsiwn unigol.

$$24x+28-4x^{2}=\frac{-64\left(x+1\right)^{2}}{4^{2}}+64\left(1-\frac{1}{2}\right)x+32$$

Tynnu \frac{1}{2} o 1 i gael \frac{1}{2}.

$$24x+28-4x^{2}=\frac{-64\left(x+1\right)^{2}}{4^{2}}+64\times \frac{1}{2}x+32$$

Lluosi 64 a \frac{1}{2} i gael 32.

$$24x+28-4x^{2}=\frac{-64\left(x+1\right)^{2}}{4^{2}}+32x+32$$

I ychwanegu neu dynnu mynegiannau, rhaid i chi eu ehangu i wneud eu enwaduron yr un fath. Lluoswch 32x+32 â \frac{4^{2}}{4^{2}}.

$$24x+28-4x^{2}=\frac{-64\left(x+1\right)^{2}}{4^{2}}+\frac{\left(32x+32\right)\times 4^{2}}{4^{2}}$$

Gan fod gan \frac{-64\left(x+1\right)^{2}}{4^{2}} a \frac{\left(32x+32\right)\times 4^{2}}{4^{2}} yr un dynodydd, adiwch nhw drwy adio eu rhifiaduron.

$$24x+28-4x^{2}=\frac{-64\left(x+1\right)^{2}+\left(32x+32\right)\times 4^{2}}{4^{2}}$$

Gwnewch y gwaith lluosi yn -64\left(x+1\right)^{2}+\left(32x+32\right)\times 4^{2}.

$$24x+28-4x^{2}=\frac{-64x^{2}-128x-64+512x+512}{4^{2}}$$

Cyfuno termau tebyg yn -64x^{2}-128x-64+512x+512.

$$24x+28-4x^{2}=\frac{-64x^{2}+384x+448}{4^{2}}$$

Cyfrifo 4 i bŵer 2 a chael 16.

$$24x+28-4x^{2}=\frac{-64x^{2}+384x+448}{16}$$

Rhannu pob term -64x^{2}+384x+448 â 16 i gael -4x^{2}+24x+28.

$$24x+28-4x^{2}=-4x^{2}+24x+28$$

Ychwanegu 4x^{2} at y ddwy ochr.

$$24x+28-4x^{2}+4x^{2}=24x+28$$

Cyfuno -4x^{2} a 4x^{2} i gael 0.

$$24x+28=24x+28$$

Tynnu 24x o'r ddwy ochr.

$$24x+28-24x=28$$

Cyfuno 24x a -24x i gael 0.

$$28=28$$

Cymharu 28 gyda 28.

$$\text{true}$$

Mae hyn yn wir ar gyfer unrhyw x.

$$x\in \mathrm{C}$$

Datrys ar gyfer x

$$x\in \mathrm{R}$$

Camau Ateb Byr

Lluoswch ddwy ochr yr hafaliad wrth 64, lluoswm cyffredin lleiaf 4,64,2.

$$16\left(x+2\right)-\left(2\left(1-x\right)\right)^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

Defnyddio’r briodwedd ddosbarthu i luosi 16 â x+2.

$$16x+32-\left(2\left(1-x\right)\right)^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

Defnyddio’r briodwedd ddosbarthu i luosi 2 â 1-x.

$$16x+32-\left(2-2x\right)^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

Defnyddio'r theorem binomaidd \left(a-b\right)^{2}=a^{2}-2ab+b^{2} i ehangu'r \left(2-2x\right)^{2}.

$$16x+32-\left(4-8x+4x^{2}\right)=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

I ddod o hyd i wrthwyneb 4-8x+4x^{2}, dewch o hyd i wrthwyneb pob term.

$$16x+32-4+8x-4x^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

Tynnu 4 o 32 i gael 28.

$$16x+28+8x-4x^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

Cyfuno 16x a 8x i gael 24x.

$$24x+28-4x^{2}=-64\times \left(\frac{x+1}{4}\right)^{2}+64\left(1-\frac{1}{2}\right)x+32$$

I godi \frac{x+1}{4} i bŵer, codwch y rhifiadur a'r enwadur i bŵer ac yna rhannwch nhw.

$$24x+28-4x^{2}=-64\times \frac{\left(x+1\right)^{2}}{4^{2}}+64\left(1-\frac{1}{2}\right)x+32$$

Mynegwch -64\times \frac{\left(x+1\right)^{2}}{4^{2}} fel ffracsiwn unigol.

$$24x+28-4x^{2}=\frac{-64\left(x+1\right)^{2}}{4^{2}}+64\left(1-\frac{1}{2}\right)x+32$$

Tynnu \frac{1}{2} o 1 i gael \frac{1}{2}.

$$24x+28-4x^{2}=\frac{-64\left(x+1\right)^{2}}{4^{2}}+64\times \frac{1}{2}x+32$$

Lluosi 64 a \frac{1}{2} i gael 32.

$$24x+28-4x^{2}=\frac{-64\left(x+1\right)^{2}}{4^{2}}+32x+32$$

I ychwanegu neu dynnu mynegiannau, rhaid i chi eu ehangu i wneud eu enwaduron yr un fath. Lluoswch 32x+32 â \frac{4^{2}}{4^{2}}.

$$24x+28-4x^{2}=\frac{-64\left(x+1\right)^{2}}{4^{2}}+\frac{\left(32x+32\right)\times 4^{2}}{4^{2}}$$

Gan fod gan \frac{-64\left(x+1\right)^{2}}{4^{2}} a \frac{\left(32x+32\right)\times 4^{2}}{4^{2}} yr un dynodydd, adiwch nhw drwy adio eu rhifiaduron.

$$24x+28-4x^{2}=\frac{-64\left(x+1\right)^{2}+\left(32x+32\right)\times 4^{2}}{4^{2}}$$

Gwnewch y gwaith lluosi yn -64\left(x+1\right)^{2}+\left(32x+32\right)\times 4^{2}.

$$24x+28-4x^{2}=\frac{-64x^{2}-128x-64+512x+512}{4^{2}}$$

Cyfuno termau tebyg yn -64x^{2}-128x-64+512x+512.

$$24x+28-4x^{2}=\frac{-64x^{2}+384x+448}{4^{2}}$$

Cyfrifo 4 i bŵer 2 a chael 16.

$$24x+28-4x^{2}=\frac{-64x^{2}+384x+448}{16}$$

Rhannu pob term -64x^{2}+384x+448 â 16 i gael -4x^{2}+24x+28.

$$24x+28-4x^{2}=-4x^{2}+24x+28$$

Ychwanegu 4x^{2} at y ddwy ochr.

$$24x+28-4x^{2}+4x^{2}=24x+28$$

Cyfuno -4x^{2} a 4x^{2} i gael 0.

$$24x+28=24x+28$$

Tynnu 24x o'r ddwy ochr.

$$24x+28-24x=28$$

Cyfuno 24x a -24x i gael 0.

$$28=28$$

Cymharu 28 gyda 28.

$$\text{true}$$

Mae hyn yn wir ar gyfer unrhyw x.

$$x\in \mathrm{R}$$