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một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?
a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)
\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)
\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-5}{2x\left(x-5\right)}\)
c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)
\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)
b: \(=\dfrac{4x\left(x-1\right)\left(x+1\right)}{6x\left(x-1\right)}=\dfrac{2\left(x+1\right)}{3}\)
c: \(=\dfrac{\left(5-x-1\right)\left(5+x+1\right)}{\left(x+6\right)^2}=\dfrac{\left(4-x\right)\left(x+6\right)}{\left(x+6\right)^2}=\dfrac{4-x}{x+6}\)
d: \(=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\dfrac{x+3}{x+2}\)
d) \(4x^2-9-x\left(2x-3\right)=0\)
\(\Leftrightarrow4x^2-9-2x^2+3x=0\)
\(\Leftrightarrow2x^2+3x-9=0\)
\(\Delta=3^2-4.2.\left(-9\right)=9+72=81\)
Vậy pt có 2 nghiệm phân biệt
\(x_1=\frac{-3+\sqrt{81}}{4}=\frac{-3}{2}\);\(x_1=\frac{-3-\sqrt{81}}{4}=-3\)
e) \(x^3+5x^2+9x=-45\)
\(\Leftrightarrow x^3+5x^2+9x+45=0\)
\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)
\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)
a: \(=\dfrac{6x^3+13x^2-5x}{2x+5}=\dfrac{6x^3+15x^2-2x^2-5x}{2x+5}=3x^2-x\)
b: \(=\dfrac{x^4-6x^3+12x^2-14x+3}{x^2-4x+1}\)
\(=\dfrac{x^4-4x^3+x^2-2x^3+8x^2-2x+3x^2-12x+3}{x^2-4x+1}\)
\(=x^2-2x+3\)
d: \(=\dfrac{\left(x+y\right)^2}{x+y}=x+y\)
e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
f: Ta có: \(x^3-6x^2+12x-19=0\)
\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)
\(\Leftrightarrow\left(x-2\right)^3=11\)
hay \(x=\sqrt[3]{11}+2\)
a) 4x(3x-7)-6(2x2-5x+1)=12
=>4x.3x-4x.7-6.2x2-6.(-5x)-6.1=12
=>12x2-28x-12x2+30x-6=12
=>2x-6 =12
=>2x =12+6
=>2x =18
=>x =18:2
=>x =6
b)(5x+3)(4x-1)+(10x-7)(-2x+3)=27
=>5x.4x-5x.1+3.4x+3.(-1)+10x.(-2x)+10x.3-7.-(2x)-7.3=27
=>20x2-5x+12x-3-20x2+30x+14x-21=27
=>39x-36 =27
=>39x =27+36
=>39x =63
=>x =63:39
=>x =21/13
c) (8x-5)(3x+2)-(12x+7)(2x-1)=17
=>8x.3x+8x.2-5.3x-5.2-12x.2x-12x.(-1)+7.2x+7.(-1)=17
=>24x2+16x-15x-10-24x2+12x+14x-7=17
=>27x-17 =17
=>27x =17+17
=>27x =34
=>x =34:27
=>x =34/27
d) (5x+9)(6x-1)-(2x-3)(15x+1)=-190
=>30x2-5x+63x-9 - 30x2-2x-45x-3=-190
=>11x-12 =-190
=>11x =-190+12
=>11x =-178
=>x = -178:11
=>x =-178/11
1, \(\frac{x^2+2x+1}{2x^2-2}=\frac{\left(x+1\right)^2}{2\left(x^2-1\right)}=\frac{\left(x+1\right)^2}{2\left(x+1\right)\left(x-1\right)}=\frac{x+1}{2\left(x-1\right)}\)= \(\frac{x+1}{2x-2}\)
2 \(\frac{x^2-6x+9}{5x^2-45}=\frac{\left(x-3\right)^2}{5\left(x^2-9\right)}=\frac{\left(x-3\right)^2}{5\left(x-3\right)\left(x+3\right)}=\frac{x-3}{5x+15}\)
3 \(\frac{x^2-12x+36}{2x^2-4x}=\frac{\left(x-6\right)^2}{2x\left(x-2\right)}\)
4 \(\frac{x^2-10x+25}{2x^2-50}=\frac{\left(x-5\right)^2}{2\left(x^2-25\right)}=\frac{\left(x-5\right)^2}{2\left(x-5\right)\left(x+5\right)}=\frac{x-5}{2x+10}\)