Question

[Min_T^T]

Mọi người giải hộ mình với ạ !

bài phương trình tích:

(x-1)(3x-6)=0

(2x+5)(1-3x)=0

(x+1)(2x-3)(3x-5)=0

6(x-2)(x-4)(1-7x)=0

(x+1)²(x+2)=0

(3x-2)²(x+1)(x-2)=0

(5-x)²(3x-1)=0

(14-2x)²(3-x)(2x-4)=0

(5x-6)²(x+2)(x+10)=0

(3x-3)³(x+4)=0

Giúp mình với nhé cảm ơn nhiều ^^

Answer 1

[Nguyễn Lê Phước Thịnh]

a) Ta có: \(\left(x-1\right)\left(3x-6\right)=0\)

\(\Leftrightarrow\left(x-1\right)\cdot3\cdot\left(x-2\right)=0\)

Vì 3≠0

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

Vậy: x∈{1;2}

b) Ta có: \(\left(2x+5\right)\left(1-3x\right)=0\)

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

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

c) Ta có: \(\left(x+1\right)\left(2x-3\right)\left(3x-5\right)=0\)

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

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

d) Ta có: \(6\left(x-2\right)\left(x-4\right)\left(1-7x\right)=0\)

Vì 6≠0

nên \(\left[{}\begin{matrix}x-2=0\\x-4=0\\1-7x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\\7x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\\x=\frac{1}{7}\end{matrix}\right.\)

Vậy: \(x\in\left\{2;4;\frac{1}{7}\right\}\)

e) Ta có: \(\left(x+1\right)^2\cdot\left(x+2\right)=0\)

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

Vậy: x∈{-1;-2}

f) Ta có: \(\left(3x-2\right)^2\cdot\left(x+1\right)\cdot\left(x-2\right)=0\)

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

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

g) Ta có: \(\left(5-x\right)^2\left(3x-1\right)=0\)

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

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

h) Ta có: \(\left(14-2x\right)^2\cdot\left(3-x\right)\cdot\left(2x-4\right)=0\)

\(\Leftrightarrow4\left(7-x\right)^2\cdot\left(3-x\right)\cdot2\cdot\left(x-2\right)=0\)

\(\Leftrightarrow8\cdot\left(7-x\right)^2\cdot\left(3-x\right)\cdot\left(x-2\right)=0\)

Vì 8≠0

nên \(\left[{}\begin{matrix}\left(7-x\right)^2=0\\3-x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7-x=0\\x=3\\x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\\x=2\end{matrix}\right.\)

Vậy: x∈{7;3;2}

i) Ta có: \(\left(5x-6\right)^2\cdot\left(x+2\right)\cdot\left(x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(5x-6\right)^2=0\\x+2=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x-6=0\\x=-2\\x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=6\\x=-2\\x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{6}{5}\\x=-2\\x=-10\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{6}{5};-2;-10\right\}\)

j) Ta có: \(\left(3x-3\right)^3\cdot\left(x+4\right)=0\)

\(\Leftrightarrow27\cdot\left(x-1\right)^3\cdot\left(x+4\right)=0\)

Vì 27≠0

nên \(\left[{}\begin{matrix}\left(x-1\right)^3=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

Vậy: x∈{1;-4}

Answer 2

[Nguyễn Lê Phước Thịnh]

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