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\(A=x^2-4x-x\left(x-4\right)-15\)
\(=x^2-4x-x^2+4x-15=-15\) => đpcm
\(B=5x\left(x^2-x\right)-x^2\left(5x-5\right)-13\)
\(=5x^3-5x^2-5x^3+5x^2-13=-13\) => đpcm
\(C=-3x\left(x-5\right)+3\left(x^2-4x\right)-3x+7\)
\(=-3x^2+15x+3x^2-12x-3x+7=7\) => đpcm
\(D=7\left(x^2-5x+3\right)-x\left(7x-35\right)-14\)
\(=7x^2-35x+21-7x^2+35x-14=7\) => đpcm
\(E=4x\left(x^2-7+2\right)-4\left(x^3-7x+2x-5\right)\)
\(=4x^3-20x-4x^3+20x+20=20\) => đpcm
\(H=x\left(5x-3\right)-x^2\left(x-1\right)+x\left(x^2-6x\right)-10+3x\)
\(=5x^2-3x-x^3+x^2+x^3-6x^2-10x+3x=-10\) => đpcm
Áp dụng công thức: \(A\left(x\right).B\left(x\right)=0\Leftrightarrow\left[{}\begin{matrix}A\left(x\right)=0\\B\left(x\right)=0\end{matrix}\right.\)
a) \(PT\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}\)
b) \(PT\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.\)
Vậy: \(S=\left\{3;20\right\}\)
c) Vì \(x^2+1\ge1>0\forall x\)
\(\Rightarrow4x+2=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)
d) \(PT\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}\)
a: =>3x-2=0 hoặc 4x+5=0
=>x=2/3 hoặc x=-5/4
b: =>(x-3)(x+20)=0
=>x=3 hoặc x=-20
c: =>4x+2=0
hay x=-1/2
d: =>2x+7=0 hoặc x-5=0 hoặc 5x+1=0
=>x=-7/2 hoặc x=5 hoặc x=-1/5
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
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`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
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`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
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`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
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`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
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`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
a: =>|x-7|=3-2x
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)
b: =>|2x-3|=4x+9
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)
c: =>3x+5=2-5x hoặc 3x+5=5x-2
=>8x=-3 hoặc -2x=-7
=>x=-3/8 hoặc x=7/2
\(a,\left(3x-2\right)\left(4x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\4x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x=2\\4x=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{5}{4}\end{cases}}\)
Vậy ............
\(b,\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2,3x-6,9=0\\0,1x+2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}2,3x=6,9\\0,1x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-20\end{cases}}\)
Vậy ...........
\(c,\left(4x+2\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x+2=0\\x^2+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=-2\\x^2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-0,5\\x\in\varnothing\end{cases}}\)
Vậy .........................
\(d,\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}2x+7=0\\x-5=0\\5x+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x=-7\\x=5\\5x=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{7}{2}\\x=5\\x=-\frac{1}{5}\end{cases}}\)
Vậy ...............
a) \(\left(3x-2\right)\left(4x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\4x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{5}{4}\end{cases}}\)
b) \(\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2,3x-6,9=0\\0,1x+2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-20\end{cases}}\)
c) \(\left(4x+2\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x+2=0\\x^2+1=0\end{cases}}\)
\(\Leftrightarrow x=-\frac{1}{2}\) ( do \(x^2+1\ge1>0\forall x\) )
d) \(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+7=0\\x-5=0\end{cases}hoặc5x+1=0}\)
\(\Leftrightarrow x\in\left\{-\frac{7}{2},5,-\frac{1}{5}\right\}\)
