Tìm x

\(x^2=36\)

\(x^2=6^2=\left(-6\right)^2\)

\(\Rightarrow x=\pm6\)

Vậy \(x=\pm6\).

\(3x^3=81\)

\(x^3=81\div3\)

\(x^3=27\)

\(x^3=3^3\)

\(\Rightarrow x=3\)

Vậy \(x=3\).

\(\left(4x\right)^2=64\)

\(\left(4x\right)^2=8^2=\left(-8\right)^2\)

\(\Rightarrow\orbr{\begin{cases}4x=8\\4x=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

Vậy \(x=\pm2\).

\(\left(x-2\right)^2=121\)

\(\left(x-2\right)^2=11^2=\left(-11\right)^2\)

\(\Rightarrow\orbr{\begin{cases}x-2=11\\x-2=-11\end{cases}}\Rightarrow\orbr{\begin{cases}x=13\\x=-9\end{cases}}\)

Vậy \(x\in\left\{13;-9\right\}\).

\(a,x^2=36\)

\(\Rightarrow x^2=6^2\)

\(\Rightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)

\(b,3x^3=81\)

\(\Rightarrow x^3=81:3\)

\(\Rightarrow x^3=27\)

\(\Rightarrow x^3=3^3\)

\(\Rightarrow x=3\)

\(c,\left(4x\right)^2=64\)

\(\Rightarrow\left(4x\right)^2=8^2\)

\(\Rightarrow\orbr{\begin{cases}4x=8\\4x=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

\(d,\left(x-2\right)^2=121\)

\(\Rightarrow\left(x-2\right)^2=11^2\)

\(\Rightarrow\orbr{\begin{cases}x-2=11\\x-2=-11\end{cases}}\Rightarrow\orbr{\begin{cases}x=13\\x=-9\end{cases}}\)

Học tốt

X-\(\frac{3}{2}\)+X-\(\frac{5}{6}\)=\(-\frac{1}{3}\)

➜2X=\(-\frac{1}{3}\)+\(\frac{3}{2}+\frac{5}{6}\)

➜ 2X=2

➜X = 1

Vậy....................

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

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

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

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

\(\Rightarrow\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)

a, Giải phương trình \(x^2-x-2=0\)

\(=''-1''^2-4\times1\times''-2''=1+8\) lớn hơn \(0\)

\(\sqrt{\Delta}=\sqrt{9}=3\)

\(\Rightarrow x_1=-1;x_2=2\)

b, Vẽ đồ thị bảng số 

- Hàm số \(y=x^2\) 

- Hàm số \(y=x+2\)

+ Cho \(x=0\Rightarrow2\) được điểm A '' 0,2 ''

+ Cho \(x=2\Rightarrow y=0\) được điềm '' -2 ; 0 '' 

Đồi thị hàm số

\(\left(x-1\right)^3+\left(2x-1\right)^3=\left(3x-2\right)^3\)

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

\(\left(3x-2\right).\left(-3\right)\left(2x^2-3x+1\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{2}\\x=\frac{2}{3}\end{matrix}\right.\)

Vậy ....

 Vì \(x_2\)là nghiệm của phương trình

=> \(x_2^2-5x_2+3=0\)

=> \(x_2+1=x^2_2-4x_2+4=\left(x_2-2\right)^2\)

Theo viet ta có

\(\hept{\begin{cases}x_1+x_2=5\\x_1x_2_{ }=3\end{cases}}\)=> \(x_1^2+x_2^2=19\)

Khi đó

\(A=||x_1-2|-|x_2-2||\)

=> \(A^2=\left(x^2_1+x_2^2\right)-4\left(x_1+x_2\right)+8-2|\left(x_1-2\right)\left(x_2-2\right)|\)

=> \(A^2=19-4.5+8-2|3-2.5+4|=1\)

Mà A>0(đề bài)

=> A=1

Vậy A=1

Nhận thấy \(x=0\) không phải nghiệm, chia 2 vế cho \(x^2\)

\(6x^2+7x-36+\frac{7}{x}+\frac{6}{x^2}=0\)

\(\Leftrightarrow6\left(x^2+\frac{1}{x^2}\right)+7\left(x+\frac{1}{x}\right)-36=0\)

Đặt \(x+\frac{1}{x}=a\) (\(\left|a\right|\ge2\)) \(\Rightarrow x^2+\frac{1}{x^2}=a^2-2\)

\(6\left(a^2-2\right)+7a-36=0\)

\(\Leftrightarrow6a^2+7a-48=0\)

Nghiệm xấu