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LV
1
26 tháng 2 2022
a: =>5x-5+17x=1-12x-4
=>22x-5=-12x-3
=>34x=2
hay x=1/17
b: =>\(\left(x-3\right)^2-4x\left(x-3\right)=0\)
=>(x-3)(-3x-3)=0
=>x=3 hoặc x=-1
c: =>(x-4)(x-6)=0
=>x=4 hoặc x=6
10 tháng 8 2023
a: \(=\dfrac{6x^2+9x+8x+12}{2x+3}=\dfrac{3x\left(2x+3\right)+4\left(2x+3\right)}{2x+3}\)
=3x+4
b: \(=\dfrac{5x^2-2x+15x-6}{5x-2}\)
\(=\dfrac{x\left(5x-2\right)+3\left(5x-2\right)}{5x-2}=x+3\)
c: \(=\dfrac{-8x^2+20x+2x-5-10}{2x-5}=-4x+1+\dfrac{-10}{2x-5}\)
d: \(=\dfrac{14x^2-35x+2x-5}{2x-5}=\dfrac{7x\left(2x-5\right)+\left(2x-5\right)}{2x-5}\)
=7x+1
e: \(=\dfrac{2x^3+x^2+6x^2+3x+12x+6}{2x+1}\)
\(=\dfrac{x^2\left(2x+1\right)+3x\left(2x+1\right)+6\left(2x+1\right)}{2x+1}=x^2+3x+6\)
f: \(=\dfrac{x^3-2x^2+6x^2-12x+x-2}{x-2}=x^2+6x+1\)
g: \(=\dfrac{12x^3+6x^2-4x^2-2x+6x+3}{2x+1}=6x^2-2x+3\)
LV
18 tháng 1 2022
một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?
28 tháng 4 2023
a: =>17x-5x-15-2x-5=0
=>10x-20=0
=>x=2
b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)
=>11x+23=-2x-16
=>13x=-39
=>x=-3(nhận)
c: =>5x+7>=3x-3
=>2x>=-10
=>x>=-5
d: =>5(3x-1)=-2(x+1)
=>15x-5=-2x-2
=>17x=3
=>x=3/17
e: =>4x^2-1-4x^2-3x-2=0
=>-3x-3=0
=>x=-1
g: =>7x-5-8x+2-7<0
=>-x-10<0
=>x+10>0
=>x>-10
28 tháng 4 2023
a: =>17x-5x-15-2x-5=0
=>10x-20=0
=>x=2
b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)
=>11x+23=-2x-16
=>13x=-39
=>x=-3(nhận)
c: =>5x+7>=3x-3
=>2x>=-10
=>x>=-5
d: =>5(3x-1)=-2(x+1)
=>15x-5=-2x-2
=>17x=3
=>x=3/17
e: =>4x^2-1-4x^2-3x-2=0
=>-3x-3=0
=>x=-1
g: =>7x-5-8x+2-7<0
=>-x-10<0
=>x+10>0
=>x>-10
KT
1
3 tháng 8 2020
1) Ta có: \(\left(x^2-4x+4\right)\left(x^2+4x+4\right)-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)^2\cdot\left(x+2\right)^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\right]^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x^2-4\right)^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x^2-4-7x-4\right)\left(x^2-4+7x+4\right)=0\)
\(\Leftrightarrow\left(x^2-7x-8\right)\left(x^2+7x\right)=0\)
\(\Leftrightarrow x\left(x+7\right)\left(x^2-8x+x-8\right)=0\)
\(\Leftrightarrow x\left(x+7\right)\left[x\left(x-8\right)+\left(x-8\right)\right]=0\)
\(\Leftrightarrow x\left(x+7\right)\left(x-8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\\x-8=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\\x=8\\x=-1\end{matrix}\right.\)
Vậy: S={0;-7;8;-1}
2) Ta có: \(x^3-8x^2+17x-10=0\)
\(\Leftrightarrow x^3-2x^2-6x^2+12x+5x-10=0\)
\(\Leftrightarrow x^2\left(x-2\right)-6x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-6x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-x-5x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=5\end{matrix}\right.\)
Vậy: S={2;1;5}
3) Ta có: \(2x^3-5x^2-x+6=0\)
\(\Leftrightarrow2x^3-4x^2-x^2+2x-3x+6=0\)
\(\Leftrightarrow2x^2\left(x-2\right)-x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-3x+2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(2x-3\right)+\left(2x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\2x=3\\x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{3}{2}\\x=-1\end{matrix}\right.\)
Vậy: \(S=\left\{2;\frac{3}{2};-1\right\}\)
4) Ta có: \(4x^4-4x^2-3=0\)
\(\Leftrightarrow4x^4-6x^2+2x^2-3=0\)
\(\Leftrightarrow2x^2\left(2x^2-3\right)+\left(2x^2-3\right)=0\)
\(\Leftrightarrow\left(2x^2-3\right)\left(2x^2+1\right)=0\)
mà \(2x^2+1>0\forall x\in R\)
nên \(2x^2-3=0\)
\(\Leftrightarrow2x^2=3\)
\(\Leftrightarrow x^2=\frac{3}{2}\)
hay \(x=\pm\sqrt{\frac{3}{2}}\)
Vậy: \(S=\left\{\sqrt{\frac{3}{2}};-\sqrt{\frac{3}{2}}\right\}\)
(x + 3)(x - 5) - (4x + 1)(1 - 4x) = 17x2
=> x2 - 5x + 3x - 15 - 1 + 16x2 = 17x2
=> 17x2 - 2x - 16 - 17x2 = 0
=> 2x = -16
=> x = -16 : 2 = -8