\(-2\left(\sqrt{1+x}+\sqrt{1-x}\right)+7=\sqrt{\left(5-2x\right)\left(5+2x\right)}-2\sqrt{1-x^2}\)

ĐKCĐ: \(-1\le x\le1\)

\(\Leftrightarrow2\left(\sqrt{\left(1-x\right)}-1\right)\left(\sqrt{1+x}-1\right)+5-\sqrt{\left(5-2x\right)\left(5+2x\right)}=0\)

 \(\Leftrightarrow2x^2\left[\frac{2}{5+\sqrt{\left(5-2x\right)\left(5+2x\right)}}-\frac{1}{\left(\sqrt{1-x}+1\right)\left(\sqrt{1+x}+1\right)}\right]\)

Đặt: \(A=\frac{2}{5+\sqrt{\left(5-2x\right)\left(5+2x\right)}}-\frac{1}{\left(\sqrt{1-x}+1\right)\left(\sqrt{1+x}+1\right)}\)

Có: \(A\le\frac{2}{5+\sqrt{\left(5-2\right)\left(5-2\right)}}-\frac{1}{\sqrt{1-x^2}+1+\sqrt{1-x}+\sqrt{1+x}}< \frac{2}{5+3}-\frac{1}{1+1+2}=0\)

\(\Rightarrow x=0\) là nghiệm của pt

x=0. Ai giúp với
 

Mấy ché k giúp ak.

ĐK: \(\hept{\begin{cases}x^3+2x+4\ge0\\x^3-2x+4\ge0\end{cases}}\)

Đặt: \(\hept{\begin{cases}a=\sqrt{x^3+2x+4}\left(a\ge0\right)\\b=\sqrt{x^3-2x+4}\left(b\ge0\right)\end{cases}\Rightarrow\hept{\begin{cases}a^2=x^3+2x+4\\b^2=x^3-2x+4\end{cases}}\Rightarrow a^2-b^2=4x\Rightarrow x=\frac{a^2-b^2}{4}}\) 

\(pt\Leftrightarrow\left[1+\left(\frac{a^2-b^2}{4}\right)\right]a+\left[1-\left(\frac{a^2-b^2}{4}\right)\right]b=4\) 

\(\Leftrightarrow\left(4+a^2-b^2\right)a+\left(4-a^2+b^2\right)b=16\)

\(\Leftrightarrow a^3+b^3-ab^2-a^2b+4\left(a+b\right)=16\)

\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)+4\left(a+b\right)=16\)

\(\Leftrightarrow\left(a+b\right)\left(a^2-2ab+b^2\right)+4\left(a+b\right)=16\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2+4\left(a+b\right)=16\) (1)

Từ pt, ta có: \(\left(1+x\right)a-\left(1-x\right)b=4\)

\(\Leftrightarrow a+b+\left(a-b\right)x=4\) (2)

Thay (1) và (2) vào, ta có:

\(\left(a+b\right)\left(a-b\right)^2+4\left(a+b\right)=4\left[a+b+\left(a-b\right)x\right]\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2=4\left(a-b\right)x\)

\(\Leftrightarrow\left(a-b\right)\left[\left(a+b\right)\left(a-b\right)-4x\right]=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2-4x\right)=0\Leftrightarrow\orbr{\begin{cases}a=b\\a^2-b^2=4x\end{cases}}\)

Với \(a=b\) , ta có: \(\sqrt{x^3+2x+4}=\sqrt{x^3-2x+4}\Leftrightarrow x=0\left(TM\right)\)

Với \(a^2-b^2=4x\) , ta có: \(x^3+2x+4-\left(x^3-2x+4\right)=4x\)

\(\Leftrightarrow4x=0\)

\(\Rightarrow x=0\)

Vậy:.........

Lớp mấy đây, lớp 8 mà đây á

tớ ra =0 cậu k cho mình nhé

ĐKXĐ: \(x>0\)

Ta có:

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

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

\(\frac{\Leftrightarrow1}{2x\sqrt{x}}-\sqrt{x}=2\left(x-\frac{1}{x}+\frac{1}{4x^3}\right)\)

Đặt : \(\frac{1}{2x\sqrt{x}}-\sqrt{x}=a\Rightarrow a^2=x-\frac{1}{x}+\frac{1}{4x^3}\)

Khi đó pt đã cho trở thành:

\(a=2a^2\Leftrightarrow\orbr{\begin{cases}a=0\\a=\frac{1}{2}\end{cases}}\)

+) a = 0\(\Rightarrow x=\frac{1}{\sqrt{2}}\)

Tương tự

dk \(\hept{\begin{cases}x\left(3x+1\right)\ge0\\x\left(x-1\right)\ge0\end{cases}< =>\orbr{\begin{cases}x\ge1\\x\le\frac{-1}{3}\end{cases}}}\)

vì x khác 0 nên chia cả 2 vế cho \(\sqrt{x}\)ta được \(\sqrt{3x+1}-\sqrt{x-1}=2\sqrt{x}< =>\)\(\sqrt{x-1}+2\sqrt{x}-\sqrt{3x+1}=0< =>\)\(\sqrt{x-1}+\frac{4x-\left(3x+1\right)}{2\sqrt{x}+\sqrt{3x+1}}=0\)\(\sqrt{x-1}+\frac{x-1}{2\sqrt{x}+\sqrt{3x+1}}=0\)\(< =>\sqrt{x-1}\left(1+\frac{\sqrt{x-1}}{2\sqrt{x}+\sqrt{3x+1}}\right)=0< =>\sqrt{x-1}=0\) (vì biểu thức trong ngoặc luôn \(\ge1\)) <=> x-1= 0 <=> x=1 (thỏa mãn điều kiện)

b. 2 + \(\sqrt{2x-1}=x\)       ĐKXĐ: \(x\ge0,5\)

<=> \(\sqrt{2x-1}\) = x - 2

<=> 2x - 1 = (x - 2)²

<=> 2x - 1 = x² - 4x + 4

<=> -x² + 2x + 4x - 4 - 1 = 0

<=> -x² + 6x - 5 = 0

<=> -x² + 5x + x - 5 = 0

<=> -(-x² + 5x + x - 5) = 0

<=> x² - 5x - x + 5 = 0

<=> x(x - 5) - (x - 5) = 0

<=> (x - 1)(x - 5) = 0

<=> \(\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)