x-[-x+(x+3)]-[x+30-(x-2)]=0
Những câu hỏi liên quan
tìm x sao cho x^3+x^2-x+2=0; x^3-6x^2-x+30=0
Tìm x:
1, x(x-20)+x-2=0
2, 5x(x-3)-x+3=0
3, 3x(x-5)-(x-1)(2+3x)=30
4, (x+2)(x+3)-(x-2)(x+5)=0
tìm x biết
1/ x(x -2)-x-2 =0
2/ 5x(x-3)-x+3=0
3/ 3x(x-5)-(x-1)(2+3x) =30
4/ (x+2)(x+3)-(x-2)(x+5)=0
4x
\(1\) . \(x\left(x-2\right)-x+2=0\)
\(\Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\Rightarrow x=2\\x-1=0\Rightarrow x=1\end{matrix}\right.\)
2 . \(5x\left(x-3\right)-x+3=0\)
\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\5x-1=0\Rightarrow x=\dfrac{1}{5}\end{matrix}\right.\)
3 . \(3x\left(x-5\right)-\left(x-1\right)\left(2+3x\right)=30\)
\(\Leftrightarrow3x^2-15x-2x-3x^2+2+3x=30\)
\(\Leftrightarrow-14x+2=30\)
\(\Leftrightarrow-14x=28\)
\(\Rightarrow x=-2\)
4 . \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)
\(\Leftrightarrow2x+16=0\)
\(\Rightarrow x=-8\)
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Tìm x biết
a)(x+3)^2(x-2)^2=2x b)7x(x-2)=(x-2) c)8x^3-12x^2+6x-1=0
d)4x^2-9-x(2x-3)=0 e)x^3+5x^2+9x=-45 f)x^3-6x^2-x+30=0
d) \(4x^2-9-x\left(2x-3\right)=0\)
\(\Leftrightarrow4x^2-9-2x^2+3x=0\)
\(\Leftrightarrow2x^2+3x-9=0\)
\(\Delta=3^2-4.2.\left(-9\right)=9+72=81\)
Vậy pt có 2 nghiệm phân biệt
\(x_1=\frac{-3+\sqrt{81}}{4}=\frac{-3}{2}\);\(x_1=\frac{-3-\sqrt{81}}{4}=-3\)
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e) \(x^3+5x^2+9x=-45\)
\(\Leftrightarrow x^3+5x^2+9x+45=0\)
\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)
\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)
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f) \(x^3-6x^2-x+30=0\)
\(\Leftrightarrow\left(x^3-x^2-6x\right)-\left(5x^2-5x-30\right)=0\)
\(\Leftrightarrow x\left(x^2-x-6\right)-5\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2-2x+3x-6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left[x\left(x-2\right)+3\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow x\in\left\{5;-3;2\right\}\)
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Xem thêm câu trả lời
(X-30)^800=(x-30)^900
Bài 2
X^2-2^x= 0
Bài 3
2017^x=2016^x
|x+3|+|-x-3|=20
-(-1-x)+(-x)^3+x(x^2-2)=-30
2*x*x*x+3*(-x)^2*x+4*(-x)^3-5*(-x)=-10
x^2(x-2)-16x+32=0
|x+3|=|2x-9|
(2x+9)^2=(x+15)^2
|x-15|-|2x+6|=0
Tìm x thuộc Z biết
a)(x+1)(x-2)<0
b)(x+2)(x+6)>0
c)(x+1)(8-x)<0
d)(x3+1)(x3+10)(x3+12)(x3+30)<0
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Nếu thấy hay thì cho mk 1 ckkk nhé
lên lớp 8 ròi làm ddaaau có sao
1)M = [ x∈ N/30:x]
2) P = [ x∈ N/20<x<30:3]
3) Q = [ x∈ N/x.(x-7)(3x+5)=0]
Tìm x
a) (x + 3)2 + (x + 2)(5 – x) = 1
b/ (2x – 1)2 – ( x – 5)( 4x + 3) = 3
c/ 3x (x – 2) + 4x – 8 = 0
d/ 2x (3x + 5) – 18x – 30 = 0
\(a,\Leftrightarrow x^2+6x+9-x^2+3x+10=1\\ \Leftrightarrow9x=-18\Leftrightarrow x=-2\\ b,\Leftrightarrow4x^2-4x+1-4x^2+17x+15=3\\ \Leftrightarrow13x=-13\Leftrightarrow x=-1\\ c,\Leftrightarrow3x\left(x-2\right)+4\left(x-2\right)=0\\ \Leftrightarrow\left(3x+4\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=2\end{matrix}\right.\\ d,\Leftrightarrow2x\left(3x+5\right)-6\left(3x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)
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