a) chắc là nhóm lại thui để sau mk làm:v

b)\(\sqrt{\frac{x+7}{x+1}}+8=2x^2+\sqrt{2x-1}\)

Đk: tự lm nhé :v

\(pt\Leftrightarrow\sqrt{\frac{x+7}{x+1}}-\sqrt{3}-\left(\sqrt{2x-1}-\sqrt{3}\right)=2x^2-8\)

\(\Leftrightarrow\frac{\frac{x+7}{x+1}-3}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}-\frac{2x-1-3}{\sqrt{2x-1}+\sqrt{3}}=2\left(x^2-4\right)\)

\(\Leftrightarrow\frac{\frac{-2x+4}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}-\frac{2\left(x-2\right)}{\sqrt{2x-1}+\sqrt{3}}=2\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow\frac{\frac{-2\left(x-2\right)}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}-\frac{2\left(x-2\right)}{\sqrt{2x-1}+\sqrt{3}}-2\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(\frac{\frac{-2}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}-\frac{2}{\sqrt{2x-1}+\sqrt{3}}-2\left(x+2\right)\right)=0\)

Dễ thấy: \(\frac{\frac{-2}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}-\frac{2}{\sqrt{2x-1}+\sqrt{3}}-2\left(x+2\right)< 0\)

\(\Rightarrow x-2=0\Rightarrow x=2\)

ban tra loi nhanh giup mk nhe

Nhanh lên bạn ơi , mai mình phải nộp rồi . 

ĐK:\(-1\le x\le1\)

\(\sqrt{\frac{1+2x\sqrt{1-x^2}}{2}}=1-2x^2\)

\(\Leftrightarrow\frac{1+2x\sqrt{1-x^2}}{2}=4x^4-4x^2+1\)

\(\Leftrightarrow\frac{2x\sqrt{1-x^2}}{2}=\frac{8x^4-8x^2+1}{2}\)

\(\Leftrightarrow2x\sqrt{1-x^2}=8x^4-8x^2+1\)

\(\Leftrightarrow4x^2\left(1-x^2\right)=64x^8-128x^6+80x^4-16x^2+1\)

\(\Leftrightarrow-\left(2x^2-1\right)^2\left(16x^4-16x^2+1\right)=0\)

Suy ra \(2x^2-1=0\)  hoặc \(16x^4-16x^2+1=0\)

Suy ra \(x=-\frac{1}{\sqrt{2}}\) hoặc \(16\left(x^2-\frac{1}{2}\right)^2-3=0\Rightarrow x=\frac{\sqrt{12}-2}{\sqrt{32}}\) (thỏa)

ĐK:\(-1\le x\le1\)

\(\sqrt{\frac{1+2x\sqrt{1-x^2}}{2}}=1-2x^2\)

\(\Leftrightarrow\frac{1+2x\sqrt{1-x^2}}{2}=4x^4-4x^2+1\)

\(\Leftrightarrow\frac{2x\sqrt{1-x^2}}{2}=\frac{8x^4-8x^2+1}{2}\)

\(\Leftrightarrow2x\sqrt{1-x^2}=8x^4-8x^2+1\)

\(\Leftrightarrow4x^2\left(1-x^2\right)=64x^8-128x^6+80x^4-16x^2+1\)

\(\Leftrightarrow-\left(2x^2-1\right)^2\left(16x^4-16x^2+1\right)=0\)

Suy ra \(2x^2-1=0\)  hoặc \(16x^4-16x^2+1=0\)

Suy ra \(x=-\frac{1}{\sqrt{2}}\) hoặc \(16\left(x^2-\frac{1}{2}\right)^2-3=0\Rightarrow x=\frac{\sqrt{12}-2}{\sqrt{32}}\) (thỏa)

\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\left(x\ne1\right)\)

\(\Leftrightarrow\frac{1}{x-1}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{4}{x^2+x+1}=0\)

\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{4x-4}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Leftrightarrow\frac{x^2+x+1+2x^2-5-4x+4}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Leftrightarrow\frac{3x^2-3x}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Leftrightarrow\frac{3x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Rightarrow3x=0\)

=> x=0 (tmđk)
Vậy x=0

Dk: x\(\ge0\)

lien hop

\(\Leftrightarrow\sqrt{x+3}-\sqrt{x}=1\)

\(\Leftrightarrow\sqrt{x+3}=2\Rightarrow x=1\)

B​ạn có thể giải thích rõ hộ mình dc k???

oh ok mình hiểu r thanks =)))

TỰ LÀM ĐI

\(2x+\left|x-\frac{1}{2}\right|=2\)

Điều kiện x \(\ge\frac{1}{4}\)

Đặt a = \(\sqrt{x-\frac{1}{4}}\)(a \(\ge0\))

=> x = a² + \(\frac{1}{4}\)

=> PT <=> 2a² + \(\frac{1}{2}\)+ \(\sqrt{a^2+\frac{1}{4}+a}\)= 2

<=> \(\sqrt{a^2+\frac{1}{4}+a}\)= \(\frac{3}{2}-2a\)

<=> a² + 0,25 + a = 4a⁴ + 2,25 - 6a²

<=> 4a⁴ - 7a² - a + 2 = 0

<=> (a + 1)(2a - 1)(2a² - a - 2) = 0

<=> a = 0,5

<=> x = 0,5