\(\left|x+19\right|+\left|x+5\right|+\left|x+2011\right|=4x\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
VT>=0 suy ra 4x>=0
suy ra x>=0
..................................................................................................
Do : VP ≥ 0
=> VT ≥ 0
=> 4x ≥ 0
=> x ≥ 0
nên Phương trình trên có dạng :
x + 2 + x + 9 + x + 2011 = 4x
<=> 3x + 2022 = 4x
<=> x = 2022 ( thỏa mãn )
KL....
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x+1\right)\left(2x-1\right)\left(-x+2\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+1=0\\2x-1=0\\-x+2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-1\\x=\frac{1}{2}\\x=2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{-1;\frac{1}{2};2\right\}\)
b) \(\left(2x-1\right)\left(3x+2\right)\left(4x-5\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}2x-1=0\\3x+2=0\\4x-5=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=\frac{1}{2}\\x=-\frac{2}{3}\\x=\frac{5}{4}\\x=7\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{\frac{1}{2};-\frac{2}{3};\frac{5}{4};7\right\}\)
c) \(x^2-6x+11=0\)
\(\Leftrightarrow x^2-6x+9+2=0\)
\(\Leftrightarrow\left(x-3\right)^2+2=0\) (vô lí)
Vậy phương trình vô nghiệm
d) \(\left(x^2+2x+3\right)\left(x^2-25\right)\left(x+19\right)=0\)
\(\Leftrightarrow\left(x^2+2x+1+2\right)\left(x+5\right)\left(x-5\right)\left(x+19\right)=0\)
\(\Leftrightarrow\left[\left(x+1\right)^2+2\right]\left(x+5\right)\left(x-5\right)\left(x+19\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+5=0\\x-5=0\\x+19=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-5\\x=5\\x=-19\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{\pm5;-19\right\}\)
a,b,d dễ mà bạn tự làm
c,x2-6x+11=0<=> x2-6x+9+2=0
<=>(x-3)2=-2(vô lý)
vậy pt vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^3-6x^2-9x+14=0\)
\(\Leftrightarrow x^3-8x^2+2x^2+7x-16x+14=0\)
\(\Leftrightarrow\left(x^3-8x^2+7x\right)+\left(2x^2-16x+14\right)=0\)
\(\Leftrightarrow x\left(x^2-8x+7\right)+2\left(x^2-8x+7\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-8x+7\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-7x-x+7\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[x\left(x-7\right)-\left(x-7\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x-7\right)=0\)
\(\Leftrightarrow x\in\left\{-2;1;7\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a)
$x^3-6x^2-9x+14=0$
$\Leftrightarrow x^3-x^2-5x^2+5x-14x+14=0$
$\Leftrightarrow x^2(x-1)-5x(x-1)-14(x-1)=0$
$\Leftrightarrow (x-1)(x^2-5x-14)=0$
$\Leftrightarrow (x-1)(x^2-7x+2x-14)=0$
$\Leftrightarrow (x-1)[x(x-7)+2(x-7)]=0$
$\Leftrightarrow (x-1)(x+2)(x-7)=0$
$\Rightarrow x=1; x=-2$ hoặc $x=7$
b)
Bạn tham khảo tại đây:
Câu hỏi của Lương Đức Hưng - Toán lớp 8 | Học trực tuyến
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\left(x^2-16\right)\left(\dfrac{x}{4}-\dfrac{4x+5}{3}\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\left(\dfrac{3x-16x-20}{12}\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\cdot\left(-13x-20\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\-13x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\-13x=20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=\dfrac{-20}{13}\end{matrix}\right.\)
Vậy: \(x\in\left\{4;-4;\dfrac{-20}{13}\right\}\)
b) Ta có: \(\left(4x-1\right)\left(x+5\right)=x^2-25\)
\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(4x-1-x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-5;\dfrac{-4}{3}\right\}\)
c) Ta có: \(x\left(x+3\right)^3-\dfrac{x}{4}\cdot\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\cdot\left[x\left(x+3\right)^2-\dfrac{1}{4}x\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left[x\left(x^2+6x+9\right)-\dfrac{1}{4}x\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^3+6x^2+9x-\dfrac{1}{4}x\right)=0\)
\(\Leftrightarrow\left(x+3\right)\cdot x\cdot\left(x^2+6x+\dfrac{35}{4}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x^2+6x+9-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left[\left(x+3\right)^2-\dfrac{1}{4}\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x+3-\dfrac{1}{2}\right)\left(x+3+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x+\dfrac{5}{2}\right)\left(x+\dfrac{7}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+\dfrac{5}{2}=0\\x+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-3;-\dfrac{5}{2};-\dfrac{7}{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=46\)
\(\Leftrightarrow16x^2-9-16x^2+40x-25=46\)
\(\Leftrightarrow40x=46+9+25=80\)
\(\Leftrightarrow x=2\)
b) \(\left(x+1\right)^3+2x-\left(x-1\right)^3-3\left[\left(x+1\right)^2+\left(x-1\right)^2\right]+5=0\)
\(=x^3+3x^2+3x+1+2x-x^3+3x^2-3x+1-3\left(x^2+2x+1+x^2-2x+1\right)+5=0\)
\(=6x^2+2x+2-3\left(2x^2+2\right)+5=0\)
\(\Leftrightarrow6x^2+2x+2-6x^2-6+5=0\)
\(\Leftrightarrow2x=-2+6-5=-1\)
\(\Leftrightarrow x=\frac{1}{2}\)