Question

Giải các phương trình sau:

a) (x-1)(3x-7)=(x-1)(x+3)

b) (3x+1)(x-2)=(x-2)(x+1)

c) (x-3)^2 +2x-6=0

d)(2x-1)^2=9

e) x(x-1)=3(x-1)

f)x^2(x-3)=4(x-3)

g) x(x-5)=2(x-5)

Answer 1

a) \(\left(x-1\right).\left(3x-7\right)=\left(x-1\right).\left(x+3\right)\)

\(\Leftrightarrow\left(x-1\right).\left(3x-7\right)-\left(x-1\right).\left(x+3\right)=0\)

\(\Leftrightarrow\left(x-1\right).\left(3x-7-x-3\right)=0\)

\(\Leftrightarrow\left(x-1\right).\left(2x-10\right)=0\)

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

Vì \(2\ne0.\)

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

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{1;5\right\}.\)

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

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

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

\(\Leftrightarrow\left(x-2\right).2x=0\)

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

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;0\right\}.\)

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

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

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

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

\(\Leftrightarrow\left(x-3\right).\left(x-1\right)=0\)

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

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;1\right\}.\)

d) \(\left(2x-1\right)^2=9\)

\(\Leftrightarrow\left(2x-1\right)^2=\left(\pm3\right)^2\)

\(\Leftrightarrow2x-1=\pm3.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4:2\\x=\left(-2\right):2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;-1\right\}.\)

Chúc bạn học tốt!

Answer 2

a) Ta có: \(\left(x-1\right)\left(3x-7\right)=\left(x-1\right)\left(x+3\right)\)

⇔\(\left(x-1\right)\left(3x-7\right)-\left(x-1\right)\left(x+3\right)=0\)

⇔\(\left(x-1\right)\left(3x-7-x-3\right)=0\)

⇔\(\left(x-1\right)\left(2x-10\right)=0\)

⇔\(\left(x-1\right)\cdot2\cdot\left(x-5\right)=0\)

Vì 2≠0

nên \(\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

Vậy: x∈{1;5}

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

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

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

\(\Leftrightarrow\left(x-2\right)\cdot2x=0\)

Vì 2≠0

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

Vậy: x∈{0;2}

c) Ta có: \(\left(x-3\right)^2+2x-6=0\)

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

⇔\(\left(x-3\right)\left(x-3+2\right)=0\)

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

⇔\(\left[{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy: x∈{3;1}

d) Ta có: \(\left(2x-1\right)^2=9\)

⇔\(\left(2x-1\right)^2-9=0\)

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

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

⇔\(2\left(x-2\right)\cdot2\cdot\left(x+1\right)=0\)

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

Vì 4≠0

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

Vậy: x∈{2;-1}

e) Ta có: \(x\left(x-1\right)=3\left(x-1\right)\)

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

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

⇔\(\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy: x∈{1;3}

f) Ta có: \(x^2\left(x-3\right)=4\left(x-3\right)\)

⇔\(x^2\left(x-3\right)-4\left(x-3\right)=0\)

⇔\(\left(x-3\right)\left(x^2-4\right)=0\)

⇔\(\left(x-3\right)\left(x-2\right)\left(x+2\right)=0\)

⇔\(\left[{}\begin{matrix}x-3=0\\x-2=0\\x+2=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=3\\x=2\\x=-2\end{matrix}\right.\)

Vậy: x∈{3;2;-2}

g) Ta có: \(x\left(x-5\right)=2\left(x-5\right)\)

⇔\(x\left(x-5\right)-2\left(x-5\right)=0\)

⇔\(\left(x-5\right)\left(x-2\right)=0\)

⇔\(\left[{}\begin{matrix}x-5=0\\x-2=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)

Vậy: x∈{5;2}