2x+6x=24
tìm x
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`6^(2x-1)=24`
`(6^2)^x = 144`
`36^x = 144`
`=>` Không có giá trị nguyên của `x` thỏa mãn.
(x+2)(x+3)(x+4)(x+5)=24
=>\(\left(x^2+7x+10\right)\left(x^2+7x+12\right)=24\)
=>\(\left(x^2+7x+11\right)^2-1=24\)
=>\(\left(x^2+7x+11\right)^2-25=0\)
=>\(\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)=0\)
=>\(\left(x^2+7x+6\right)\left(x^2+7x+16\right)=0\)
=>\(x^2+7x+6=0\)
=>(x+1)(x+6)=0
=>x=-1 hoặc x=-6
\(a,\Rightarrow 2x^2-10x-3x-2x^2=26\\ \Rightarrow -13x=26\\ \Rightarrow x=-2\\ b, \Rightarrow -2x^2+3x+3-3x-3+2x^2-x=18\\ \Rightarrow -x=18\Rightarrow x=-18\)
Sửa đề: \(P=x^4+2x^2+2\left(x^2+1\right)\left(x^2+6x-1\right)+\left(x^2-6x+1\right)^2\)
\(=x^4+2x^2+1+2\cdot\left(x^2+1\right)\left(x^2+6x-1\right)+\left(x^2-6x+1\right)^2-1\)
\(=\left(x^2+1\right)^2+2\cdot\left(x^2+1\right)\left(x^2+6x-1\right)+\left(x^2+6x-1\right)^2-1\)
\(=\left(x^2+1+x^2+6x-1\right)^2-1=\left(2x^2+6x\right)^2-1\)
\(=99^2-1=\left(99-1\right)\left(99+1\right)=98\cdot100=9800\)
Sửa đề: \(P=x^4+2x^2+2\left(x^2+1\right)\left(x^2+6x-1\right)+\left(x^2-6x+1\right)^2\)
\(=x^4+2x^2+1+2\cdot\left(x^2+1\right)\left(x^2+6x-1\right)+\left(x^2-6x+1\right)^2-1\)
\(=\left(x^2+1\right)^2+2\cdot\left(x^2+1\right)\left(x^2+6x-1\right)+\left(x^2+6x-1\right)^2-1\)
\(=\left(x^2+1+x^2+6x-1\right)^2-1=\left(2x^2+6x\right)^2-1\)
\(=99^2-1=\left(99-1\right)\left(99+1\right)=98\cdot100=9800\)
Bài 1
A= (x-2)(2x-1)-2x(x+3)=2x2-x-4x+2-2x2-6x=-11x+2
Bài 1:
a) \(A=\left(x-2\right)\left(2x-1\right)-2x\left(x+3\right)\)
\(A=2x^2-x-4x+2-2x^2-6x\)
\(A=-11x+2\)
b) \(B=\left(3x-2\right)\left(2x+1\right)-\left(6x-1\right)\left(x+2\right)\)
\(B=6x^2+3x-4x-2-6x^2-12x+x+2\)
\(B=-12x\)
c) \(C=6x\left(2x+3\right)-\left(4x-1\right)\left(3x-2\right)\)
\(C=12x^2+18x-12x^2+8x+3x-2\)
\(C=29x-2\)
d) \(D=\left(2x+3\right)\left(5x-2\right)+\left(x+4\right)\left(2x-1\right)-6x\left(2x-3\right)\)
\(D=10x^2-4x+15x-6+2x^2-x+8x-4-12x^2+18x\)
\(D=36x-10\)
Ta có:
2x+6x=24
=>x(2+6)=24
=>8x=24
=>x=24:8
=>x=3
Vậy x=3
2x+6x=24 => (2+6)x=24 => 8x=24 => x= 24:8 => x=3