Tìm x , biết :
1, 17-x=7-6x
2, -12+x=5x+20
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A
1
DT
tìm x,biết
5-x-16=40+x
4x-10=15-x
15-x=4x-5
x-15=6+4x
-12+x=5x-20
7x-4=20+3x
5x-7=-21-2x
x+15=20-4x
17-x=7-6x
6
23 tháng 9 2021
c: Ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)
\(\Leftrightarrow3x^2+26x=0\)
\(\Leftrightarrow x\left(3x+26\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)
23 tháng 9 2021
\(a,\Leftrightarrow x^2+8x+16-x^3-12x^2=16\\ \Leftrightarrow x^3+11x^2-8x=0\\ \Leftrightarrow x\left(x^2+11x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+11x-8=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=121+32=153\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11-3\sqrt{17}}{2}\\x=\dfrac{-11+3\sqrt{17}}{2}\end{matrix}\right.\\ S=\left\{0;\dfrac{-11-3\sqrt{17}}{2};\dfrac{-11+3\sqrt{17}}{2}\right\}\)
\(c,\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\\ \Leftrightarrow3x^2+26x=0\\ \Leftrightarrow x\left(3x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\\ d,\Leftrightarrow x^3-6x^2+12x-8-x^3-125-6x^2=11\\ \Leftrightarrow-12x^2+12x-144=0\\ \Leftrightarrow x^2-x+12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)
Y
2
25 tháng 7 2019
(Phần a mình lấy vế phải bằng 0 nha ^^)
a,
\(\left(5x-1\right)^2-\left(5x-4\right)\left(5x+4\right)+7=0\\ \Leftrightarrow25x^2-10x+1-\left(25x^2-16\right)+7=0\\ \Leftrightarrow25x^2-10x+1-25x^2+16+7=0\\ \Leftrightarrow-10x+24=0\\ \Leftrightarrow x=2,4\)
b,
\(5x^2+4xy+4y^2+4x+1=0\left(1\right)\\ \Leftrightarrow4x^2+4x+1+x^2+4xy+4y^2=0\\ \Leftrightarrow\left(2x+1\right)^2+\left(x+2y\right)^2=0\left(1a\right)\)
Do \(VT\ge0\) với \(\forall x,y\in R\) nên:
\(\left(1a\right)\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\x+2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{1}{2}\\y=\frac{1}{4}\end{matrix}\right.\)
c,
\(\left(x+2\right)^3-x\left(x-1\right)\left(x+1\right)=6x^2+21\\ \Leftrightarrow x^3+6x^2+12x+8-x\left(x^2-1\right)-6x^2-21=0\\ \Leftrightarrow x^3+12x+8-x^3+x-21=0\\ \Leftrightarrow13x-13=0\\ \Leftrightarrow x=1\)
Chúc bạn học tốt nha
.
25 tháng 7 2019
\(b)5x^2 + 4xy + 4y^2 + 4x + 1 = 0\)
\(\Leftrightarrow\) \(4x^2 + 4x + 1 + x^2 + 4xy + 4y^2 = 0\)
\(\Leftrightarrow\)\((2x + 1)^2 + (x + 2y)^2 = 0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\x+2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{1}{2}\\y=\frac{1}{4}\end{matrix}\right.\)
\(c)(x+2)^3-x(x-1)(x+1)=6x^2+21\)
\(\Leftrightarrow x^3+6x^2+12x+8-x\left(x^2-1\right)=6x^2+21\\ \Leftrightarrow13x+8=21\\ \Leftrightarrow13x=21-8\\ \Leftrightarrow13x=13\\ \Leftrightarrow x=1\)
Y
1
31 tháng 1 2019
1,-12 + x = 5x - 20
-12 + 20 = 5x - x
8 = 4x
x = 2
2, 7x - 4 = 20 + 3x
7x - 3x = 20 + 4
4x = 24
x = 6
3 , 5x - 7 = -21 - 2x
5x + 2x = -21 + 7
7x = -14
x = -14 : 7 = -2
4 , x + 15 = 20 - 4x
x + 4x = 20 - 15
5x = 5
x = 5 : 5 = 1
5, 17 - x = 7 - 6x
-x + 6x = 7 - 17
5x = -10
x = -10 : 5 = -2