5x-5/2x+5=2x-11/x-5 eching

$$\frac { 5 x - 5 } { 2 x + 5 } = \frac { 2 x - 11 } { x - 5 }$$

x uchun yechish

$$x=8$$

$$x=10$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati -\frac{5}{2},5 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(x-5\right)\left(2x+5\right) ga, 2x+5,x-5 ning eng kichik karralisiga ko‘paytiring.

$$\left(x-5\right)\left(5x-5\right)=\left(2x+5\right)\left(2x-11\right)$$

x-5 ga 5x-5 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$5x^{2}-30x+25=\left(2x+5\right)\left(2x-11\right)$$

2x+5 ga 2x-11 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$5x^{2}-30x+25=4x^{2}-12x-55$$

Ikkala tarafdan 4x^{2} ni ayirish.

$$5x^{2}-30x+25-4x^{2}=-12x-55$$

x^{2} ni olish uchun 5x^{2} va -4x^{2} ni birlashtirish.

$$x^{2}-30x+25=-12x-55$$

12x ni ikki tarafga qo’shing.

$$x^{2}-30x+25+12x=-55$$

-18x ni olish uchun -30x va 12x ni birlashtirish.

$$x^{2}-18x+25=-55$$

55 ni ikki tarafga qo’shing.

$$x^{2}-18x+25+55=0$$

80 olish uchun 25 va 55'ni qo'shing.

$$x^{2}-18x+80=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -18 ni b va 80 ni c bilan almashtiring.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 80}}{2}$$

-18 kvadratini chiqarish.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 80}}{2}$$

-4 ni 80 marotabaga ko'paytirish.

$$x=\frac{-\left(-18\right)±\sqrt{324-320}}{2}$$

324 ni -320 ga qo'shish.

$$x=\frac{-\left(-18\right)±\sqrt{4}}{2}$$

4 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-18\right)±2}{2}$$

-18 ning teskarisi 18 ga teng.

$$x=\frac{18±2}{2}$$

x=\frac{18±2}{2} tenglamasini yeching, bunda ± musbat. 18 ni 2 ga qo'shish.

$$x=\frac{20}{2}$$

20 ni 2 ga bo'lish.

$$x=10$$

x=\frac{18±2}{2} tenglamasini yeching, bunda ± manfiy. 18 dan 2 ni ayirish.

$$x=\frac{16}{2}$$

16 ni 2 ga bo'lish.

$$x=8$$

Tenglama yechildi.

$$x=10 x=8$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati -\frac{5}{2},5 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(x-5\right)\left(2x+5\right) ga, 2x+5,x-5 ning eng kichik karralisiga ko‘paytiring.

$$\left(x-5\right)\left(5x-5\right)=\left(2x+5\right)\left(2x-11\right)$$

x-5 ga 5x-5 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$5x^{2}-30x+25=\left(2x+5\right)\left(2x-11\right)$$

2x+5 ga 2x-11 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$5x^{2}-30x+25=4x^{2}-12x-55$$

Ikkala tarafdan 4x^{2} ni ayirish.

$$5x^{2}-30x+25-4x^{2}=-12x-55$$

x^{2} ni olish uchun 5x^{2} va -4x^{2} ni birlashtirish.

$$x^{2}-30x+25=-12x-55$$

12x ni ikki tarafga qo’shing.

$$x^{2}-30x+25+12x=-55$$

-18x ni olish uchun -30x va 12x ni birlashtirish.

$$x^{2}-18x+25=-55$$

Ikkala tarafdan 25 ni ayirish.

$$x^{2}-18x=-55-25$$

-80 olish uchun -55 dan 25 ni ayirish.

$$x^{2}-18x=-80$$

-18 ni bo‘lish, x shartining koeffitsienti, 2 ga -9 olish uchun. Keyin, -9 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-18x+\left(-9\right)^{2}=-80+\left(-9\right)^{2}$$

-9 kvadratini chiqarish.

$$x^{2}-18x+81=-80+81$$

-80 ni 81 ga qo'shish.

$$x^{2}-18x+81=1$$

x^{2}-18x+81 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-9\right)^{2}=1$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-9\right)^{2}}=\sqrt{1}$$

Qisqartirish.

$$x-9=1 x-9=-1$$

9 ni tenglamaning ikkala tarafiga qo'shish.

$$x=10 x=8$$