2x + 2x+3 + 2x+5 = 164
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\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Leftrightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{3x-2y+4z}{3\cdot\dfrac{3}{2}-2\cdot\dfrac{4}{3}+4\cdot\dfrac{5}{4}}=\dfrac{-164}{\dfrac{41}{6}}=-24\\ \Leftrightarrow\left\{{}\begin{matrix}x=-24\cdot\dfrac{3}{2}=-36\\y=-24\cdot\dfrac{4}{3}=-32\\z=-24\cdot\dfrac{5}{4}=-30\end{matrix}\right.\)
g: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
đây là 1 hằng đẳng thức luôn
\(=\left(2x-3-2x-5\right)^2=\left(-8\right)^2=64\)
\(\Leftrightarrow\left(2x+3-2x+5\right)^2=x^2+6x+64\)
=>x^2+6x=0
=>x(x+6)=0
=>x=0 hoặc x=-6
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
a: Sửa đề: (5-2x)(5+2x)+2x(x+3)=4-2x^2
=>25-4x^2+2x^2+6x=4-2x^2
=>6x+25=4
=>6x=-21
=>x=-7/2
b: (3x-2)(-2x)+5x^2=-x(x-3)
=>-6x^2+4x+5x^2=-x^2+3x
=>4x=3x
=>x=0
c: =>7-(4x^2-9)=x^2+8x+16
=>7-4x^2+9-x^2-8x-16=0
=>-5x^2-8x=0
=>5x^2+8x=0
=>x(5x+8)=0
=>x=0 hoặc x=-8/5
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
Ta có: \(\left(2x+3\right)^2+\left(2x+5\right)^2-2\left(2x+3\right)\left(2x+5\right)\)
\(=\left(2x+3-2x-5\right)^2\)
\(=\left(-2\right)^2=4\)
2x + 2x+3 + 2x+5 = 164
2x.1 + 2x.23 + 2x.25 = 164
2x.1 + 2x.8 + 2x.32 = 164
2x.(1 + 8 + 32) = 164
2x.41 = 164
2x = 164 : 41
2x = 4
2x = 22
=> x = 2
Vậy x = 2
2x + 2x+3 + 2x+5 = 164
2x . 1+ 2x . 23 + 2x . 25 = 164
2x . ( 1 + 23 + 25 ) = 164
2x . 41 = 164
2x = 164 : 41
2x = 4
2x = 22
=> x= 2