\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)

\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=17\)

\(\Rightarrow x^3-25x-x^3-8=17\)

\(\Rightarrow25x=-25\Rightarrow x=-1\)

\(\left(x-3\right)\left(2x+8\right)\ge0\)

Th1: \(\hept{\begin{cases}x-3\ge0\\2x+8\ge0\end{cases}\Rightarrow}\hept{\begin{cases}x\ge3\\x\ge-4\end{cases}\Rightarrow}x\ge3\)

Th2: \(\hept{\begin{cases}x-3\le0\\2x+8\le0\end{cases}\Rightarrow}\hept{\begin{cases}x\le3\\x\le-4\end{cases}\Rightarrow}x\le-4\)

nhưng vì sao bạn lại ra 2 TH

\(\left(x-3\right).\left(2x+8\right)\ge0\)

\(\Rightarrow\hept{\begin{cases}x-3\ge0\\2x+8\ge0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x\ge3\\2x\ge-8\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x\ge3\\x\ge-4\end{cases}}\)

\(\Rightarrow x\ge3\)

Vậy \(x\ge3\)

(x+1)^3-x(x-2)^2+x-1=0

⇔x^3+3x^2+3x+1-x(x^2-4x+4)+x-1=0

⇔x^3+3x^2+3x+1-x^3+4x^2-4x+x-1=0

⇔7x^2=0

⇔x^2=0

⇔x=0

Vậy x=0

b,(x-2)^3-x^2(x-6)=4

⇔x^3-6x^2+12x-8-x^3+6x^2=4

⇔12x-8=4

⇔12x=12

⇔x=1

Vậy x=1

trả lời

a, x=1

chúc bn 

học tốt

\(a,(x+1)^3-x(x-2)^2+x-1=0\)

\(\Rightarrow x^3+3x^2+3x+1-x^3-4x^2+4x+x-1=0\)

\(\Rightarrow(x^3-x^3)+(3x^2-4x^2)+(3x+4x)+1+x-1=0\)

\(\Rightarrow-x^2+7x+1+x-1=0\)

\(\Rightarrow-x^2+7x+1+x=1\)

\(\Rightarrow-x^2+7x+x+1=1\)

\(\Rightarrow-x^2+8x+1=1\)

\(\Rightarrow-x^2+8x=0\)

\(\Rightarrow-x(x+8)=0\)

\(\Rightarrow\orbr{\begin{cases}-x=0\\x+8=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-8(ktm)\end{cases}}\)

Vậy x = 0

a) ${}^{}-25=0$

${}^{}=0+25=25$

${}^{}={}^{}={}^{}$

$\Rightarrow [\begin{array}{c}2x-1=5\\ 2x-1=-5\end{array}\Rightarrow [\begin{array}{c}2x=6\\ 2x=-4\end{array}\Rightarrow [\begin{array}{c}x=3\\ x=-2\end{array}$

b) $8{}^{}-50x=0$

$2x(4{}^{}$

câu b thiếu bn ơi

ok

a) ${}^{}-25=0$

${}^{}=0+25=25$

${}^{}={}^{}={}^{}$

$\Rightarrow [\begin{array}{c}2x-1=5\\ 2x-1=-5\end{array}\Rightarrow [\begin{array}{c}2x=6\\ 2x=-4\end{array}\Rightarrow [\begin{array}{c}x=3\\ x=-2\end{array}$

b) $8{}^{}-50x=0$

$2x(4{}^{}$

a: Ta có: \(\left(2x-1\right)^2-25=0\)

\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

\(1964\)

phải làm rõ ràng ra

1964 nha

Đặt \(\dfrac{x}{4}=\dfrac{y}{5}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=4k\\y=5k\end{matrix}\right.\)

Ta có: \(x^2-3y^2=-59\)

\(\Leftrightarrow16k^2-3\cdot25k^2=-59\)

\(\Leftrightarrow k^2=1\)

Trường hợp 1: k=1

\(\Leftrightarrow\left\{{}\begin{matrix}x=4k=4\\y=5k=5\end{matrix}\right.\)

Trường hợp 2: k=-1

\(\Leftrightarrow\left\{{}\begin{matrix}x=4k=-4\\y=5k=-5\end{matrix}\right.\)