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Ta có: \(2x\left(x-5\right)-x\left(3+2x\right)=26\\ < =>2x^2-10x-3x-2x^2=26\\ < =>-13x=26\\ =>x=\dfrac{26}{-13}=-2\)
2x (x - 5) - x (3 + 2x) = 26
\(\Leftrightarrow\) 2x2 - 10x - 3x - 2x2 = 26
\(\Rightarrow\) -13x = 26
\(\Rightarrow\) x = -2
Vậy x = -2
\(a,2x\left(x-5\right)-x\left(2x+3\right)=26\)
\(\Leftrightarrow2x^2-10x-2x^2-3x=26\)
\(\Leftrightarrow-13x=26\)
\(\Leftrightarrow x=-2\)
\(b,\left(3x^2-x+1\right)\left(x-1\right)+x^2\left(4-3x\right)=\frac{5}{2}\)
\(\Leftrightarrow3x^3-3x^2-x^2+x+x-1+4x^2-3x^3=\frac{5}{2}\)
\(\Leftrightarrow2x=\frac{7}{2}\)
\(\Leftrightarrow x=\frac{7}{4}\)
2x( x-5) - x( 3+2x) = 26
=> 2x^2 - 10x - 3x - 2x^2 = 26
=> -13x= 26
=> x= -2
2x(x-5)-x(3+2x)=26
2x^2-10x-3x-2x^2=26
-10x-3x=26
-x(10+3)=26
-x.13=26
-x=2
=>x=-2
\(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)
\(\Leftrightarrow-13x=26\)
\(\Leftrightarrow x=26:-13\)
\(\Leftrightarrow x=-2\)
Vậy \(x=-2\)
a. \(2x\left(x-5\right)-x\left(2x+3\right)=26\Rightarrow2x^2-10x-2x^2-3x=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
b. \(\left(3y^2-y+1\right)\left(y-1\right)+y^2\left(4-3y\right)=\frac{5}{2}\)
\(\Rightarrow3y^3-3y^2-y^2+y+y-1+4y^2-3y^3=\frac{5}{2}\)\(\Rightarrow2y=\frac{7}{2}\Rightarrow y=\frac{7}{4}\)
c. \(2x^2+3\left(x+1\right)\left(x-1\right)=5x^2+5x\Rightarrow5x^2-3=5x^2+5x\)
\(\Rightarrow x=-\frac{3}{5}\)
\(3x^3-\frac{3}{2}x^2-x^3-\frac{1}{2}x+\frac{1}{2}x+2=2x^3-\frac{3}{2}x^2+2\)
\(2x^2-10x-3x-2x^2=26\)
-13x=26
x=-2
\(2x^2-10x-3x-2x^2=26\)
\(2x^2-13x-2x^2=26\)
\(-13x=26\)
\(x=-2\)
\(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Leftrightarrow\left(2x^2-10x\right)-\left(3x+2x^2\right)=26\)
\(\Leftrightarrow-13x=26\)
\(\Leftrightarrow x=-2\)