(2x-5)3+1=-26
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`1,[(-3).3+5]-26=-2x-3`
`=>(-9+5)-26=-2x-3`
`=>-4-26=-2x-3`
`=>-30=-2x-3`
`=>-2x=-27`
`=>x=27/2`
Vậy `x=27/2`
`2)-[(-35)-3]=2x-2`
`=>-(-38)=2x-2`
`=>38=2x-2`
`=>2x=40`
`=>x=20`
Vậy `x=20`
1) \(\left[\left(-3\right)\cdot3+5\right]-26=-2x-3\\ \Rightarrow-9+5-26=-2x-3\\ \Rightarrow-2x=-9+5-26+3\\ \Rightarrow-2x=-27\\ \Rightarrow x=\dfrac{27}{2}\)
Vậy \(x=\dfrac{27}{2}\)
2) \(-\left[\left(-35\right)-3\right]=2x-2\\ \Rightarrow2x-2=-\left(-38\right)\\ \Rightarrow2x=38+2\\ \Rightarrow2x=40\\ \Rightarrow x=20\)
Vậy \(x=20\)
1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)
\(=x^2+4x+4+x^2-6x+9\)
\(=2x^2-2x+13\)
2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)
\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)
\(=-2x+7\)
3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)
\(=x^2-25-x^2-10x-25\)
=-10x-50
4) Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)
\(=x^2-6x+9-x^2+16\)
=-6x+25
5) Ta có: \(\left(y^2-6y+9\right)-\left(y-3\right)^2\)
\(=y^2-6y+9-y^2+6y-9\)
=0
6) Ta có: \(\left(2x+3\right)^2-\left(2x-3\right)\left(2x+3\right)\)
\(=4x^2+12x+9-4x^2+9\)
=12x+18
Bài 2:
a: \(\Leftrightarrow2x^2-10x-3x-2x^2=26\)
=>-13x=26
hay x=-2
b: \(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{5}\right\}\)
c: \(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)
hay \(x\in\left\{-5;2\right\}\)
a)\(5\left(2x-1\right)-4\left(8-3x\right)=7\)
\(\Leftrightarrow10x-5+12x-32=7\)
\(\Leftrightarrow22x-37=7\)
\(\Leftrightarrow22x=44\Rightarrow x=2\)
b)\(5x\left(x-5\right)-x\left(2x+3\right)=26\)
\(\Leftrightarrow5x^2-25x-2x^2-3x=26\)
\(\Leftrightarrow3x^2-28x-26=0\)
\(\Leftrightarrow3\left(x-\dfrac{14}{3}\right)^2-\dfrac{274}{3}=0\)
\(\Rightarrow x=\dfrac{14}{3}\pm\dfrac{\sqrt{274}}{3}\)
\(6x-15+1+26=0\)
\(6x+12=0\)
\(x=-2\)
\(\left(2x-5\right).3+1=-26\)
=> (2x - 5).3 = -27
=> 2x - 5 = -9
=> 2x = -4
=> x = -2