Ta có : 3x³ - 4³ = 17

=> 3x³ - 64 = 17

=> 3x³ = 81

=> x³ = 27

=> x³ = 3³

=> x = 3

\(3x^3-4^3=17 \) 

\(3x^3=17+64\)

\(3x^3=81\)

\(x^3=27\)

\(x=3\)

3x³ - 4³ = 17

3x³ = 17 + 64

3x³ = 81

x³ = 81 ÷ 3

x³ = 27

Mà 3³ = 27

=> x = 3

~ Chúc bạn học giỏi nhoa ~

a)  \(3\left(4-2x\right)-2\left(x+3\right)=12-7x\)

\(\Leftrightarrow\)\(12-6x-2x-6=12-7x\)

\(\Leftrightarrow\)\(6-8x=12-7x\)

\(\Leftrightarrow\)\(x=-6\)

Vậy...

b)  \(\left|16+\right|3\left(x-2\right)||-5=20\)

\(\Leftrightarrow\)\(\left|16+\right|3\left(x-2\right)||=25\)(1)

Ta thấy:  \(\left|3\left(x-2\right)\right|\ge0\)\(\Rightarrow\)\(16+\left|3\left(x-2\right)\right|>0\)

nên từ (1)   \(\Rightarrow\)  \(16+\left|3\left(x-2\right)\right|=25\)

                   \(\Leftrightarrow\)\(\left|3\left(x-2\right)\right|=9\)

                   \(\Leftrightarrow\)  \(\orbr{\begin{cases}3\left(x-2\right)=9\\3\left(x-2\right)=-9\end{cases}}\)

                  \(\Leftrightarrow\) \(\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

Vậy....

c)   \(\left|-5-3^2\right|-||3x+5|-7.2^3|=3^9:3^7\)

\(\Leftrightarrow\)\(14-||3x+5|-56|=9\)

\(\Leftrightarrow\)\(||3x+5|-56|=5\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}\left|3x+5\right|-56=5\\\left|3x+5\right|-56=-5\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}\left|3x+5\right|=61\\\left|3x+5\right|=51\end{cases}}\)

đến đây bn giải tiếp nhé

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

\(\Rightarrow x+\frac{1}{2}=0\)                                              \(\Rightarrow\frac{2}{3}-2x=0\)

\(x=\frac{-1}{2}\)                                                                         \(2x=\frac{2}{3}\)

                                                                                                  \(x=\frac{2}{3}:2\)

                                                                                                 \(x=\frac{1}{3}\)

KL: x = -1/2 hoặc x= 1/3

b) \(\left|2x-\frac{1}{3}\right|+\frac{5}{6}=\frac{11}{6}\)

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

TH1: \(2x-\frac{1}{3}=1\)

         \(2x=\frac{4}{3}\)

             \(x=\frac{2}{3}\)

TH2: \(2x-\frac{1}{3}=-1\)

\(2x=\frac{-2}{3}\)

\(x=\frac{-2}{3}:2\)

\(x=\frac{-1}{3}\)

KL: x =.........

Học tốt nhé bn!!!

\(3\left(2x-6\right)-4\left(1+2x\right)-2\left(x-4\right)=4-3\left(1+2x\right)-5\left(1-2x\right).\)

\(\Leftrightarrow6x-18-4-8x-2x+8=4-3-6x-5+10x\)

\(\Leftrightarrow-4x-14=4x-4\)

\(\Leftrightarrow-4x-4x=-4+14\)

\(\Leftrightarrow-8x=10\)

\(\Leftrightarrow x=-\frac{5}{4}\)

Ta có: \(\hept{\begin{cases}\left(\frac{1}{x}+y\right)+\left(\frac{1}{x}-y\right)=\frac{5}{8}\\\left(\frac{1}{x}+y\right)-\left(\frac{1}{x}-y\right)=-\frac{3}{8}\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{2}{x}=\frac{5}{8}\\2y=-\frac{3}{8}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{16}{5}\\y=-\frac{3}{16}\end{cases}}}\)

\(\left(x+3\right)^3-x\left(3x-1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-x\left(9x^2-6x+1\right)+8x^3-4x^2+2x+4x^2-2x+1=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3+6x^2-x+8x^3-4x^2+2x+4x^2-2x+1-28=0\)

\(\Leftrightarrow15x^2+26x=0\)

\(\Leftrightarrow15x\left(x+\frac{26}{15}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}15x=0\\x+\frac{26}{15}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{26}{15}\end{cases}}}\)

Đề đúng : Chứng minh : \(\frac{x^4+4}{x\left(x^2+2\right)-2x^2-\left(x-1\right)^2-1}=\frac{x^2+2x+2}{x-1}\)

Điều kiện : \(x\ne1\)

Phân tích : \(x^4+4=\left(x^4+4x^2+4\right)-4x^2=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)

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

\(=x^3-3x^2+4x-2=\left(x^3-3x^2+3x-1\right)+\left(x-1\right)=\left(x-1\right)^3+\left(x-1\right)\)

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

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