tìm x:
x^2-2x-120=0
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a: \(\left(2x-1\right)^2-x\left(4x-3\right)\)
\(=4x^2-4x+1-4x^2+3x\)
=-x+1
b: \(\left(x-2\right)\left(x-1\right)-\left(x-3\right)^2\)
\(=x^2-3x+2-\left(x^2-6x+9\right)\)
\(=x^2-3x+2-x^2+6x-9=3x-7\)
c: \(\left(x-2\right)^3-x\left(x-1\right)\left(x-2\right)\)
\(=x^3-6x^2+12x-8-x\left(x^2-3x+2\right)\)
\(=x^3-6x^2+12x-8-x^3+3x^2-2x\)
\(=-3x^2+10x-8\)
d: \(\frac12x\left(x-2\right)-\left(2x-3\right)^2\)
\(=\frac12x^2-x-\left(4x^2-12x+9\right)\)
\(=\frac12x^2-x-4x^2+12x-9=-\frac72x^2+11x-9\)
Giải:
\(x+x\) + 133\(^0\) + 85\(^0\) = 360\(^0\)
2\(x\) + (133\(^0\) + 85\(^0\)) = 360\(^0\) (tống bốn góc của tứ giác luôn bằng 360\(^0\))
2\(x\) + 218\(^0\) = 360\(^0\)
2\(x\) = 360\(^0\) - 218\(^0\)
2\(x\) = 142\(^0\)
\(x\) = 142\(^0\) : 2
\(x\) = 71\(^0\)
xét tứ giác `KJVT` có :
\(\hat{K}+\hat{J}+\hat{V}+\hat{T}=360^0\)
`=> 133^0 + 85^0 + x + x = 360^0`
`=> 218^0 + 2x = 360^0`
`=> 2x =360^0 - 218^0`
`=> 2x = 142^0`
`=> x = 142^0 : 2`
`=> x = 71^0`
Vậy `x = 71^0`
(-7\(x^2\).2y\(^4\).z\(^3\)) + 20\(x^2\).y\(^4\).z\(^3\)
= (-7 + 20)\(x^2y^4z^3\)
= 13\(x^2y^4z^3\)
a; -(3 - 2\(x\)).(\(x+1\))
= - 3.(\(x+1\)) + 2\(x\).(\(x+1\))
= -3\(x\) - 3 + 2\(x^2\) + 2\(x\)
= -(\(3x\) - 2\(x\)) - 3 + 2\(x^2\)
= -\(x\) - 3 + 2\(x^2\)
= 2\(x^2\) - \(x\) - 3
b; 2\(x\).(3 - \(x\)) - (\(x+1)^2\)
= 6\(x\) - 2\(x^2\) - (\(x^2\) + 2\(x\) + 1)
= 6\(x\) - 2\(x^2\) - \(x^2\) - 2\(x\) - 1
= (6\(x\) - 2\(x\)) - (2\(x^2\) + \(x^2\)) - 1
= 4\(x\) - 3\(x^2\) - 1
= - 3\(x^2\) + 4\(x\) - 1
c; (\(x+4\))(2 - \(x\)) - (\(x^2\) - 2)
= \(x\).(2 - \(x\)) + 4.(2 - \(x\)) - \(x^2\) + 2
= 2\(x\) - \(x^2\) + 8 - 4\(x\) - \(x^2\) + 2
= -(4\(x\) - 2\(x\)) - (\(x^2\) + \(x^2\)) + (8 + 2)
= - 2\(x\) - 2\(x^2\) + 10
= - \(2x\) - 2\(x^2\) + 10
= - 2\(x^2\) - 2\(x\) + 10
d; - 2.(\(x-1\)).3\(x\) - 3\(x\).(\(x+2\))
= - 6\(x^2\) + 6\(x\) - 3\(x^2\) - 6\(x\)
= - (6\(x^2\) + 3\(x^2\)) + (6\(x-6x\))
= - 9\(x^2\) + 0
= -9\(x^2\)
ĐKXĐ: x∉{2;-2}
Ta có: \(T=\frac{4}{x+2}+\frac{3}{x-2}+\frac{5x+2}{4-x^2}-\frac{x^2-2x+4}{x^3+8}\)
\(=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x+2}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\)
\(=\frac{3}{x+2}+\frac{3}{x-2}-\frac{5x+2}{\left(x-2)\left(x+2\right)\right)}=\frac{3x-6+3x+6-5x-2}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x-2}{\left(x+2\right)\left(x-2\right)}=\frac{1}{x+2}\)
\(\dfrac{x+2}{x}=\dfrac32\) \((\) điều kiện: \(x\ne0\) \()\)
\(\rArr\left(x+2\right)\cdot2=x\cdot3\)
\(\lrArr2\cdot x+4=x\cdot3\)
\(\) \(\lrArr2\cdot x-x\cdot3=-4\)
\(\lrArr-1\cdot x=-4\)
\(\lrArr x=4\)
Vậy \(x=4\)
ko nên làm
\(x^2-2x-120=0\)
=>\(x^2-12x+10x-120=0\)
=>x(x-12)+10(x-12)=0
=>(x-12)(x+10)=0
=>\(\left[\begin{array}{l}x-12=0\\ x+10=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=12\\ x=-10\end{array}\right.\)
\(x^2-2x-120=0\)
\(x^2-2x+1-121=0\)
\(\left(x-1\right)^2-11^2\) =0
\(\left(x-1+11\right)\left(x-1-11\right)=0\)
\(\left(x+10\right)\left(x-12\right)=0\)
\(\left[\begin{array}{l}x+10=0\Rightarrow x=-10\\ x-12=0\Rightarrow x=12\end{array}\right.\)
Vậy x = -10; x = 12