sqrt{x}+x=7-6 eching

$$\sqrt{ x } +x=7-6$$

x uchun yechish

$$x=\frac{3-\sqrt{5}}{2}\approx 0,381966011$$

Yechim qadamlari

Tenglamaning ikkala tarafidan x ni ayirish.

$$\sqrt{x}=7-6-x$$

1 olish uchun 7 dan 6 ni ayirish.

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

Tenglamaning ikkala taraf kvadratini chiqarish.

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

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

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

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(1-x\right)^{2} kengaytirilishi uchun ishlating.

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

Ikkala tarafdan 1 ni ayirish.

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

2x ni ikki tarafga qo’shing.

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

3x ni olish uchun x va 2x ni birlashtirish.

$$3x-1=x^{2}$$

Ikkala tarafdan x^{2} ni ayirish.

$$3x-1-x^{2}=0$$

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.

$$-x^{2}+3x-1=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -1 ni a, 3 ni b va -1 ni c bilan almashtiring.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-1\right)\left(-1\right)}}{2\left(-1\right)}$$

3 kvadratini chiqarish.

$$x=\frac{-3±\sqrt{9-4\left(-1\right)\left(-1\right)}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9+4\left(-1\right)}}{2\left(-1\right)}$$

4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9-4}}{2\left(-1\right)}$$

9 ni -4 ga qo'shish.

$$x=\frac{-3±\sqrt{5}}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{5}}{-2}$$

x=\frac{-3±\sqrt{5}}{-2} tenglamasini yeching, bunda ± musbat. -3 ni \sqrt{5} ga qo'shish.

$$x=\frac{\sqrt{5}-3}{-2}$$

-3+\sqrt{5} ni -2 ga bo'lish.

$$x=\frac{3-\sqrt{5}}{2}$$

x=\frac{-3±\sqrt{5}}{-2} tenglamasini yeching, bunda ± manfiy. -3 dan \sqrt{5} ni ayirish.

$$x=\frac{-\sqrt{5}-3}{-2}$$

-3-\sqrt{5} ni -2 ga bo'lish.

$$x=\frac{\sqrt{5}+3}{2}$$

Tenglama yechildi.

$$x=\frac{3-\sqrt{5}}{2} x=\frac{\sqrt{5}+3}{2}$$

\sqrt{x}+x=7-6 tenglamasida x uchun \frac{3-\sqrt{5}}{2} ni almashtiring.

$$\sqrt{\frac{3-\sqrt{5}}{2}}+\frac{3-\sqrt{5}}{2}=7-6$$

Qisqartirish. x=\frac{3-\sqrt{5}}{2} tenglamani qoniqtiradi.

$$1=1$$

\sqrt{x}+x=7-6 tenglamasida x uchun \frac{\sqrt{5}+3}{2} ni almashtiring.

$$\sqrt{\frac{\sqrt{5}+3}{2}}+\frac{\sqrt{5}+3}{2}=7-6$$

Qisqartirish. x=\frac{\sqrt{5}+3}{2} qiymati bu tenglamani qoniqtirmaydi.

$$2+5^{\frac{1}{2}}=1$$

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

$$x=\frac{3-\sqrt{5}}{2}$$