sqrt{x^2-1}=sqrt{2x+1} eching

$$\sqrt { x ^ { 2 } - 1 } = \sqrt { 2 x + 1 }$$

x uchun yechish (complex solution)

$$x=\sqrt{3}+1\approx 2,732050808$$

$$x=1-\sqrt{3}\approx -0,732050808$$

Yechim qadamlari

Tenglamaning ikkala taraf kvadratini chiqarish.

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

2 daraja ko‘rsatkichini \sqrt{x^{2}-1} ga hisoblang va x^{2}-1 ni qiymatni oling.

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

2 daraja ko‘rsatkichini \sqrt{2x+1} ga hisoblang va 2x+1 ni qiymatni oling.

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

Ikkala tarafdan 2x ni ayirish.

$$x^{2}-1-2x=1$$

Ikkala tarafdan 1 ni ayirish.

$$x^{2}-1-2x-1=0$$

-2 olish uchun -1 dan 1 ni ayirish.

$$x^{2}-2-2x=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$x^{2}-2x-2=0$$

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

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

-2 kvadratini chiqarish.

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

-4 ni -2 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4+8}}{2}$$

4 ni 8 ga qo'shish.

$$x=\frac{-\left(-2\right)±\sqrt{12}}{2}$$

12 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-2\right)±2\sqrt{3}}{2}$$

-2 ning teskarisi 2 ga teng.

$$x=\frac{2±2\sqrt{3}}{2}$$

x=\frac{2±2\sqrt{3}}{2} tenglamasini yeching, bunda ± musbat. 2 ni 2\sqrt{3} ga qo'shish.

$$x=\frac{2\sqrt{3}+2}{2}$$

2+2\sqrt{3} ni 2 ga bo'lish.

$$x=\sqrt{3}+1$$

x=\frac{2±2\sqrt{3}}{2} tenglamasini yeching, bunda ± manfiy. 2 dan 2\sqrt{3} ni ayirish.

$$x=\frac{2-2\sqrt{3}}{2}$$

2-2\sqrt{3} ni 2 ga bo'lish.

$$x=1-\sqrt{3}$$

Tenglama yechildi.

$$x=\sqrt{3}+1 x=1-\sqrt{3}$$

\sqrt{x^{2}-1}=\sqrt{2x+1} tenglamasida x uchun \sqrt{3}+1 ni almashtiring.

$$\sqrt{\left(\sqrt{3}+1\right)^{2}-1}=\sqrt{2\left(\sqrt{3}+1\right)+1}$$

Qisqartirish. x=\sqrt{3}+1 tenglamani qoniqtiradi.

$$\left(3+2\times 3^{\frac{1}{2}}\right)^{\frac{1}{2}}=\left(2\times 3^{\frac{1}{2}}+3\right)^{\frac{1}{2}}$$

\sqrt{x^{2}-1}=\sqrt{2x+1} tenglamasida x uchun 1-\sqrt{3} ni almashtiring.

$$\sqrt{\left(1-\sqrt{3}\right)^{2}-1}=\sqrt{2\left(1-\sqrt{3}\right)+1}$$

Qisqartirish. x=1-\sqrt{3} tenglamani qoniqtiradi.

$$i\left(-\left(3-2\times 3^{\frac{1}{2}}\right)\right)^{\frac{1}{2}}=i\left(-\left(3-2\times 3^{\frac{1}{2}}\right)\right)^{\frac{1}{2}}$$

\sqrt{x^{2}-1}=\sqrt{2x+1} boʻyicha barcha yechimlar roʻyxati.

$$x=\sqrt{3}+1 x=1-\sqrt{3}$$

x uchun yechish

$$x=\sqrt{3}+1\approx 2,732050808$$

Yechim qadamlari

Tenglamaning ikkala taraf kvadratini chiqarish.

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

2 daraja ko‘rsatkichini \sqrt{x^{2}-1} ga hisoblang va x^{2}-1 ni qiymatni oling.

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

2 daraja ko‘rsatkichini \sqrt{2x+1} ga hisoblang va 2x+1 ni qiymatni oling.

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

Ikkala tarafdan 2x ni ayirish.

$$x^{2}-1-2x=1$$

Ikkala tarafdan 1 ni ayirish.

$$x^{2}-1-2x-1=0$$

-2 olish uchun -1 dan 1 ni ayirish.

$$x^{2}-2-2x=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$x^{2}-2x-2=0$$

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

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

-2 kvadratini chiqarish.

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

-4 ni -2 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4+8}}{2}$$

4 ni 8 ga qo'shish.

$$x=\frac{-\left(-2\right)±\sqrt{12}}{2}$$

12 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-2\right)±2\sqrt{3}}{2}$$

-2 ning teskarisi 2 ga teng.

$$x=\frac{2±2\sqrt{3}}{2}$$

x=\frac{2±2\sqrt{3}}{2} tenglamasini yeching, bunda ± musbat. 2 ni 2\sqrt{3} ga qo'shish.

$$x=\frac{2\sqrt{3}+2}{2}$$

2+2\sqrt{3} ni 2 ga bo'lish.

$$x=\sqrt{3}+1$$

x=\frac{2±2\sqrt{3}}{2} tenglamasini yeching, bunda ± manfiy. 2 dan 2\sqrt{3} ni ayirish.

$$x=\frac{2-2\sqrt{3}}{2}$$

2-2\sqrt{3} ni 2 ga bo'lish.

$$x=1-\sqrt{3}$$

Tenglama yechildi.

$$x=\sqrt{3}+1 x=1-\sqrt{3}$$

\sqrt{x^{2}-1}=\sqrt{2x+1} tenglamasida x uchun \sqrt{3}+1 ni almashtiring.

$$\sqrt{\left(\sqrt{3}+1\right)^{2}-1}=\sqrt{2\left(\sqrt{3}+1\right)+1}$$

Qisqartirish. x=\sqrt{3}+1 tenglamani qoniqtiradi.

$$\left(3+2\times 3^{\frac{1}{2}}\right)^{\frac{1}{2}}=\left(2\times 3^{\frac{1}{2}}+3\right)^{\frac{1}{2}}$$

\sqrt{x^{2}-1}=\sqrt{2x+1} tenglamasida x uchun 1-\sqrt{3} ni almashtiring. \sqrt{\left(1-\sqrt{3}\right)^{2}-1} ifodasi noaniq, chunki ildiz ostidagi qiymat manfiy boʻlishi mumkin emas.

$$\sqrt{\left(1-\sqrt{3}\right)^{2}-1}=\sqrt{2\left(1-\sqrt{3}\right)+1}$$

\sqrt{x^{2}-1}=\sqrt{2x+1} tenglamasi noyob yechimga ega.

$$x=\sqrt{3}+1$$