a,x+12x+1x+12x+1=0.5x+2x+30.5x+2x+3
b,|9-7x|=5x-3
c,|10x+7|<37
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\(a,=6y\left(2x^2-3xy-5y^2\right)\\ =6y\left(2x^2+2xy-5xy-5y^2\right)\\ =6y\left(x+y\right)\left(2x-5y\right)\\ b,=5x\left(x-y\right)-10\left(x-y\right)=5\left(x-2\right)\left(x-y\right)\\ c,=\left(a-b\right)\left(a^2+ab+b^2\right)-3\left(a-b\right)\\ =\left(a-b\right)\left(a^2+ab+b^2-3\right)\\ d,=\left(a^2+3b\right)^2-1=\left(a^2+3b+1\right)\left(a^2+3b-1\right)\\ e,=\left(2x-5\right)\left(2x+5\right)-\left(2x+7\right)\left(2x-5\right)\\ =\left(2x-5\right)\left(2x+5-2x-7\right)\\ =-2\left(2x-5\right)\\ f,=x^2+5x-3x-15=\left(x+5\right)\left(x-3\right)\\ g,=x^3-x-6x-6\\ =x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\\ =\left(x+1\right)\left(x^2-x-6\right)\\ =\left(x+1\right)\left(x^2-3x+2x-6\right)\\ =\left(x+1\right)\left(x-3\right)\left(x+2\right)\\ l,=x^4+4x^2+4-4x^2\\ =\left(x^2+2\right)^2-4x^2=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\\ h,=y\left(x^2+2x+1\right)=y\left(x+1\right)^2\)
một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?
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
a) 6x(5x + 3) + 3x(1 – 10x) = 7
⇒ 30x2+18x+3x-30x2=7
⇒21x=7
⇒x=\(\dfrac{7}{21}\)
⇒x= \(\dfrac{1}{3}\)
b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44
⇒15x-63x2-15+63x + 63x2-35x+36x-20=44
⇒79x-35=44
⇒79x=44+35
⇒79x=79
⇒x=1
\(a,12x=4x-30\Leftrightarrow8x=-30\Leftrightarrow x=-\dfrac{15}{4}\)
\(b,2x-5=x-1\Leftrightarrow2x-x=-1+5\Leftrightarrow x=4\)
\(c,2-5x=5x-10\Leftrightarrow-10x=-12\Leftrightarrow x=\dfrac{6}{5}\)
\(d,9x-6=1x-5\Leftrightarrow8x=1\Leftrightarrow x=\dfrac{1}{8}\)
\(e,2x-5=2x-1\Leftrightarrow2x-2x=-1+5\Leftrightarrow0x=4\) (Vô lí)\(\Rightarrow x\in\varnothing\)
a: \(\Leftrightarrow x^2+4x+3=x^2+4x+0.5x+2\)
=>0,5x+2=3
=>0,5x=1
hay x=2
b: \(\Leftrightarrow\left|7x-9\right|=5x-3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(7x-9-5x+3\right)\left(7x-9+5x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(2x-6\right)\left(12x-12\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{3;1\right\}\)
c: =>10x+7>-37 và 10x+7<37
=>10x>-44 và 10x<30
=>-4,4<x<3
\(d,x\left(x-3\right)-7x+21=0\)
\(\Leftrightarrow x\left(x-3\right)-7\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=7\end{cases}}}\)
\(a,2x\left(x-7\right)+5x-35=0\)
\(\Leftrightarrow2x\left(x-7\right)+5\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-\frac{5}{2}\end{cases}}}\)
\(c,4x^2+12x+9=0\)
\(\Leftrightarrow4x^2+6x+6x+9=0\)
\(\Leftrightarrow2x\left(2x+3\right)+3\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow x=-\frac{3}{2}\)
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 \(\frac{x^2-10x+25}{2x^2-50}=\frac{\left(x-5\right)^2}{2\left(x^2-25\right)}=\frac{\left(x-5\right)^2}{2\left(x-5\right)\left(x+5\right)}=\frac{x-5}{2x+10}\)
I don't now
or no I don't
..................
sorry
a)\(x+12x+x+12x+1=0,5x+2x+30,5x+2x+3\)
\(\Leftrightarrow26x+1=35x+3\)
\(\Leftrightarrow26x+1-\left(35x+3\right)=0\)
\(\Leftrightarrow26x+1-35x-3=0\)
\(\Leftrightarrow-9x+\left(-2\right)=0\)
\(\Leftrightarrow-9x=2\)
\(\Leftrightarrow x=-\frac{2}{9}\)