Bài 1:

Ta có: \(\left(2x^2+x-4\right)^2-\left(2x-1\right)^2=0\)

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

\(\Leftrightarrow\left(2x^2-x-3\right)\left(2x^2+3x-5\right)=0\)

\(\Leftrightarrow\left(2x^2+2x-3x-3\right)\left(2x^2-2x+5x-5\right)=0\)

\(\Leftrightarrow\left[2x\left(x+1\right)-3\left(x+1\right)\right]\left[2x\left(x-1\right)+5\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x-3\right)\left(x-1\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x-3=0\\x-1=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x=3\\x=1\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{3}{2}\\x=1\\x=\frac{-5}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{-1;\frac{3}{2};1;\frac{-5}{2}\right\}\)

Còn 3 câu kia đâu bạn?

Cau 1. X=2

Cau 2 x= 23

Cau/3.x=14

ban co the nao giai chi tiet cho minh dc ko

Bài 3 nhé bạn đặt cái căn đầu là a ,căn sau là b 

a+b=x

ab=1

Rồi tính lần lượt a³ +b³ bằng ẩn x hết 

và mũ 4 cũng vậy rồi lấy 2 số nhân nhau .Bđ là ra 

=33X

a: \(\frac{7-x}{2}+\frac23\left(x-7\right)\left(x-3\right)=0\)

=>\(\frac{-\left(x-7\right)}{2}+\frac23\left(x-7\right)\left(x-3\right)=0\)

=>\(\left(x-7\right)\left(-\frac12+\frac23x-2\right)=0\)

=>\(\left(x-7\right)\left(\frac23x-\frac52\right)=0\)

=>\(\left[\begin{array}{l}x-7=0\\ \frac23x-\frac52=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ \frac23x=\frac52\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=\frac52:\frac23=\frac{15}{4}\end{array}\right.\)

b: \(\left(2x+7\right)^2=9\left(x+2\right)^2\)

=>\(\left(3x+6\right)^2=\left(2x+7\right)^2\)

=>\(\left(3x+6\right)^2-\left(2x+7\right)^2=0\)

=>(3x+6-2x-7)(3x+6+2x+7)=0

=>(x-1)(5x+13)=0

=>\(\left[\begin{array}{l}x-1=0\\ 5x+13=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=-\frac{13}{5}\end{array}\right.\)

\(x^2-2x+3=t\left(t\ge0\right)\)

\(pt\Leftrightarrow\frac{1}{t-1}+\frac{1}{t}=\frac{9}{2\left(t+1\right)}\)

\(\Leftrightarrow\frac{2t\left(t+1\right)}{2t\left(t^2-1\right)}+\frac{2\left(t^2-1\right)}{2t\left(t^2-1\right)}-\frac{9t\left(t-1\right)}{2t\left(t^2-1\right)}=0\)

\(\Leftrightarrow-5t^2+11t-2=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=2\end{cases}}\)