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

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
 

kuchiyose edo tensei

nhờ vào năng lực rinegan , ta có thể  đoán dc

  \(\left(\sqrt{1+x}+\sqrt{8-x}\right)^2=1+x+8-x-2\sqrt{\left(X+1\right)\left(8-x\right)}\)

vậy pt sẽ như sau

\(a,\left(\sqrt{1+x}+\sqrt{8-x}\right)^2-\sqrt{\left(1+x\right)\left(8-x\right)}=3\) " thêm bớt nếu m thông minh sẽ hiểu "

\(9+2\sqrt{\left(1+x\right)\left(8-x\right)}-\sqrt{\left(1+x\right)\left(8-x\right)}=3\)

\(\sqrt{\left(1+x\right)\left(8-x\right)}=-6\)

\(\left(1+x\right)\left(8-x\right)=36\)

đến đây m có thể tự làm

c)  \(\sqrt{x+5}=5-x^2\)

      \(x+5=\left(5-x\right)^2\)

     \(x+5=x^4-10x^2+25\)  " rồi xong pt bậc 4 :)

 \(x^4-10x^2-x+20=0\)

\(x^4=10x^2+x-20\)

\(x^4+2mx^2+m^2=10x^2+x-20+2mx^2+m^2\)

\(\left(x^2+m\right)^2=2x^2\left(5+m\right)+x+\left(m^2-20\right)\)

\(\Delta=1-8\left(5+m\right)\left(m^2-20\right)\)

\(\Delta=1-8\left(5m^2-100+m^3-20m\right)\)

\(\Delta=1-40m^2+800-8m^3+160m\)

\(\Delta=-\left(2m+9\right)\left(4m^2+2m-89\right)\)

lấy m= -9/2 , cho nhanh thay vào ta đươc

\(\left(x^2-\frac{9}{2}\right)^2=2x^2\left(5-\frac{9}{2}\right)+x+\left(\frac{9}{2}^2-20\right)\)

\(\left(x^2-\frac{9}{2}\right)^2=x^2+x+\frac{1}{4}\)

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

\(\hept{\begin{cases}x^2-\frac{9}{2}=x+\frac{1}{2}\\x^2-\frac{9}{2}=-x-\frac{1}{2}\end{cases}}\)

đến đây cậu có thể làm tiếp :)

câu B hơi gắt cần time suy nghĩ :)

\(A=\frac{\left(x^2+a\right)\left(1+a\right)+a^2x^2+1}{\left(x^2-a\right)\left(1-a\right)+a^2x^2+1}\)

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

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

\(=\frac{\left(1+a+a^2\right)\left(1+x^2\right)}{\left(1-a+a^2\right)\left(1+x^2\right)}=\frac{1+a+a^2}{1-a+a^2}\) không phụ thuộc vào x

a: \(A=\left(\dfrac{\sqrt{3}\left(x-\sqrt{3}\right)+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\right)\cdot\dfrac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\)

\(=\dfrac{x\sqrt{3}}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\cdot\dfrac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)

\(=\dfrac{1}{x-\sqrt{3}}\)

b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)

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

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

c: \(C=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}}\)

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

\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)

a)

ĐKĐB: \(\left\{\begin{matrix} 2x-1\geq 0\\ x^2+2x-5\geq 0\end{matrix}\right.\)

PT \(\Leftrightarrow 2x-1=x^2+2x-5\) (bình phương 2 vế)

\(\Leftrightarrow x^2-4=0\Leftrightarrow (x-2)(x+2)=0\Rightarrow \left[\begin{matrix} x=2\\ x=-2\end{matrix}\right.\)

Thử lại vào ĐKĐB suy ra $x=2$ là nghiệm duy nhất.

b)

ĐKĐB: \( \left\{\begin{matrix} x(x^3-3x+1)\geq 0\\ x(x^3-x)\geq 0\end{matrix}\right.\)

PT \(\Leftrightarrow x(x^3-3x+1)=x(x^3-x)\) (bình phương)

\(\Leftrightarrow x(x^3-3x+1-x^3+x)=0\)

\(\Leftrightarrow x(1-2x)=0\Rightarrow \left[\begin{matrix} x=0\\ x=\frac{1}{2}\end{matrix}\right.\)

Thử lại vào ĐKĐB thấy $x=0$ là nghiệm duy nhất

e)

ĐKXĐ: \(x\geq \frac{5}{3}\)

PT \(\Rightarrow (\sqrt{x+2}-\sqrt{2x-3})^2=3x-5\) (bình phương 2 vế)

\(\Leftrightarrow 3x-1-2\sqrt{(x+2)(2x-3)}=3x-5\)

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

\(\Leftrightarrow 4=(x+2)(2x-3)\)

\(\Leftrightarrow 2x^2+x-10=0\)

\(\Leftrightarrow (x-2)(2x+5)=0\Rightarrow \left[\begin{matrix} x=2\\ x=\frac{-5}{2}\end{matrix}\right.\)

Kết hợp với ĐKXĐ suy ra $x=2$

f) Bạn xem lại đề.