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\(\text{a) (5x+2)(x-7)=0}\)
\(\Leftrightarrow\orbr{\begin{cases}5x+2=0\\x-7=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{5}\\x=7\end{cases}}\)
Vậy ...
#Thảo Vy#
a) (5x - 1)(2x + 1) = (5x -1)(x + 3)
<=> (5x - 1)(2x + 1) - (5x -1)(x + 3) = 0
<=> (5x - 1)(2x + 1 - x - 3) = 0
<=> (5x - 1)(x - 2) = 0
<=> \(\orbr{\begin{cases}5x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0,2\\x=2\end{cases}}\)
Vậy x = 0,2 ; x = 2 là nghiệm phương trình
b) x3 - 5x2 - 3x + 15 = 0
<=> x2(x - 5) - 3(x - 5) = 0
<=> (x2 - 3)(x - 5) = 0
<=> \(\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\left(x-5\right)=0\)
<=> \(x-\sqrt{3}=0\text{ hoặc }x+\sqrt{3}=0\text{ hoặc }x-5=0\)
<=> \(x=\sqrt{3}\text{hoặc }x=-\sqrt{3}\text{hoặc }x=5\)
Vậy \(x\in\left\{\sqrt{3};\sqrt{-3};5\right\}\)là giá trị cần tìm
c) (x - 3)2 - (5 - 2x)2 = 0
<=> (x - 3 + 5 - 2x)(x - 3 - 5 + 2x) = 0
<=> (-x + 2)(3x - 8) = 0
<=> \(\orbr{\begin{cases}-x+2=0\\3x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{8}{3}\end{cases}}\)
Vậy tập nghiệm phương trình \(S=\left\{2;\frac{8}{3}\right\}\)
d) x3 + 4x2 + 4x = 0
<=> x(x2 + 4x + 4) = 0
<=> x(x + 2)2 = 0
<=> \(\orbr{\begin{cases}x=0\\\left(x+2\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
Vậy tập nghiệm phương trình S = \(\left\{0;-2\right\}\)
a) Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow\left(x^2-x\right)-\left(2x-2\right)=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{1;2\right\}\)
b) Ta có: \(-x^2+5x-6=0\)
\(\Leftrightarrow-\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow-\left(x^2-2x-3x+6\right)=0\)
\(\Leftrightarrow-\left[\left(x^2-2x\right)-\left(3x-6\right)\right]=0\)
\(\Leftrightarrow-\left[x\left(x-2\right)-3\left(x-2\right)\right]=0\)
\(\Leftrightarrow-\left[\left(x-2\right)\left(x-3\right)\right]=0\)
\(\Leftrightarrow-\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy: x∈{2;3}
c) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-10x-2x+5=0\)
⇔(4x2-10x)-(2x-5)=0
\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{2};\frac{5}{2}\right\}\)
d) Ta có: \(2x^2+5x+3=0\)
\(\Leftrightarrow2x^2+2x+3x+3=0\)
\(\Leftrightarrow\left(2x^2+2x\right)+\left(3x+3\right)=0\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x+3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;\frac{-3}{2}\right\}\)
e) Ta có: \(x^3+2x^2-x-2=0\)
\(\Leftrightarrow\left(x^3+2x^2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow x^2\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\\x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\\x=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;1;-1\right\}\)
g) Ta có: \(\left(3x-1\right)^2-5\left(2x+1\right)^2+\left(6x-3\right)\left(2x+1\right)=\left(x-1\right)^2\)
\(\Leftrightarrow9x^2-6x+1-20x^2-20x-5+12x^2-3-x^2+2x-1=0\)
\(\Leftrightarrow-24x-8=0\)
\(\Leftrightarrow-8\left(3x+1\right)=0\)
⇔3x+1=0
\(\Leftrightarrow3x=-1\)
\(\Leftrightarrow x=-\frac{1}{3}\)
Vậy: \(x=-\frac{1}{3}\)
h) \(2x^3-7x^2+7x-2=0\)
\(\Leftrightarrow2x^3-4x^2-3x^2+6x+x-2=0\)
\(\Leftrightarrow2x^2\left(x-2\right)-3x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-2x-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[2x\left(x-1\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy S = {2; 1; \(\frac{1}{2}\)}
i) \(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=0\)
\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\\left(x+\frac{1}{2}\right)^2=\frac{-23}{4}\left(loai\right)\end{matrix}\right.\)
Vậy S = {1;-2}
Sử dụng phương pháp phân tích thành nhân tử
( có thể nhẩm nghiệm =casio rồi tách)
mk làm VD 1 cái
mấy cái còn lại tương tự
\(x^2-3x+2=x^2-x-2x+2=0\)
\(x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(x-2\right)=0\)
=> x=1 hoặc x=2
- Kudo -
a) x2 - 3x + 2 = 0
<=> (x - 2)(x - 1) = 0
<=> x - 2 = 0 hoặc x - 1 = 0
<=> x = 2 hoặc x = 1
b) x2 + 5x + 6 =0
<=> (x + 2)(x + 3) = 0
<=> x + 2 = 0 hoặc x + 3 = 0
<=> x = -2 hoặc x = -3
c) x2 - 4x + 3 = 0
<=> (x - 1)(x - 3) = 0
<=> x - 1 = 0 hoặc x - 3 = 0
<=> x = 1 hoặc x = 3
d) x2 + 2x - 3 = 0
<=> (x - 1)(x + 3) = 0
<=> x - 1 = 0 hoặc x + 3 = 0
<=> x = 1 hoặc x = -3
e) x2 - 2x = 0
<=> x(x - 2) = 0
<=> x = 0 hoặc x - 2 = 0
<=> x = 0 hoặc x = 2
a) \(x^2-3x+2=0\)
Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: x∈{1;2}
b) \(x^2+5x+6=0\)
Ta có: \(x^2+5x+6=0\)
\(\Leftrightarrow x^2+2x+3x+6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
Vậy: x∈{-2;-3}
c) \(x^2-4x+3=0\)
Ta có: \(x^2-4x+3=0\)
\(\Leftrightarrow x^2-x-3x+3=0\)
\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy: x∈{1;3}
d) \(x^2+2x-3=0\)
Ta có: \(x^2+2x-3=0\)
\(\Leftrightarrow x^2+3x-x-3=0\)
\(\Leftrightarrow x\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
Vậy: x∈{-3;1}
e) \(x^2-2x=0\)
Ta có: \(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy: x∈{0;2}
xíu giải...............