tìm x biết: |x|+|x-1|+|x-2|+11=6x-1
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a) \(\left(x+3\right)\left(x+1\right)-x\left(x-5\right)=11\)
\(\Leftrightarrow x^2+x+3x+3-x^2+5x=11\)
\(\Leftrightarrow9x+3=11\)
\(\Leftrightarrow9x=11-3\)
\(\Leftrightarrow9x=8\)
\(\Leftrightarrow x=\dfrac{8}{9}\)
b) \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(\Leftrightarrow\left(8x-24x^2+2-6x\right)+\left(24x^2-60x-4x+10\right)=-50\)
\(\Leftrightarrow2x-24x^2+2+24x^2-64x+10=-50\)
\(\Leftrightarrow-62x+12=-50\)
\(\Leftrightarrow-62x=-50-12\)
\(\Leftrightarrow-62x=-62\)
\(\Leftrightarrow x=\dfrac{-62}{-62}\)
\(\Leftrightarrow x=1\)
a) \(\left(x+3\right)\left(x+1\right)-x\left(x-5\right)=11\)
\(x^2+x+3x+3-x^2+5x=11\)
\(x+8x+3=11\)
\(x+8x=8\)
\(x\left(8+1\right)=8\)
\(x=\dfrac{8}{9}\)
b) \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(8x-24x^2+2-6x+24x^2-60x-4x+10=-50\)
\(-62x+12=-50\)
\(-62x=-62\)
\(x=1\)
x^2 + 6x + 11 = 0
x^2 + 6x + 9 + 2 = 0
(x+3)^2 + 2 = 0
Mà (x+3)^2 + 2 > 0 với mọi x
=> pt vô ngiệm
Vậy .........
Hok tốt
x2 + 6x + 11 = 0
x . (x + 6) = 11
* x = 11
* x + 6 = 11
x = 11 - 6
x = 5
Vậy x = 11 hoặc x = 5
\(x^2+6x+11=0\)
\(x^2+2.x.3+9+2=0\)
\(x^2+2.x.3+3^2=-2\)
\(\left(x+3\right)^2=-2\)
Vì \(\left(x+3\right)^2\ge0\forall x\)
Mà\(-2< 0\)
\(\Rightarrow x\in\varnothing\)
\(x^2+6x+11=0\)
\(x\left(x+6\right)=-11\)
Ta có bảng:
x | 1 | -1 | 11 | -11 |
x+6 | -11 | 11 | -1 | 1 |
x | -17 | 5 | -7 | -5 |
Vậy..
hok tốt!!
1: Ta có: \(4x^2-36=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
2: Ta có: \(\left(x-1\right)^2+x\left(4-x\right)=11\)
\(\Leftrightarrow x^2-2x+1+4x-x^2=11\)
\(\Leftrightarrow2x=10\)
hay x=5
\(A=x^2-2x+10\)
\(A=\left(x^2-2x+1\right)+9\)
\(A=\left(x-1\right)^2+9\)
Mà \(\left(x-1\right)^2\ge0\)
\(\Rightarrow A\ge9\)
Dấu "=" xảy ra khi :
\(x-1=0\Leftrightarrow x=1\)
Vậy Min A = 9 khi x = 1
\(B=x^2-5x-7\)
\(B=\left(x^2-5x+\frac{25}{4}\right)-\frac{53}{4}\)
\(B=\left(x-\frac{5}{2}\right)^2-\frac{53}{4}\)
Mà \(\left(x-\frac{5}{2}\right)^2\ge0\)
\(\Rightarrow B\ge-\frac{53}{4}\)
Dấu "=" xảy ra khi :
\(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)
Vậy \(B_{Min}=-\frac{53}{4}\Leftrightarrow x=\frac{5}{2}\)
\((x-2)^3-(x+5)(x^2-5x+25)+6x^2=11\\\Leftrightarrow (x-2)^3-(x+5)(x^2-5.x+5^2)+6x^2=11 \\\Leftrightarrow x^3-6x^2+12x-8 -(x^3+5^3)+6x^2-11=0 \\\Leftrightarrow 12x-144=0 \\\Leftrightarrow x=12\)
Vậy \(x=12\).
(x−2)3−(x+5)(x2−5x+25)+6x2=11
=>(x−2)3−(x+5)(x2−5.x+52)+6x2=11
=>x3−6x2+12x−8−(x3+53)+6x2−11=0
=>12x−144=0
=>x=12(x−2)3−(x+5)(x2−5x+25)+6x2=11
=>(x−2)3−(x+5)(x2−5.x+52)+6x2=11
=>x3−6x2+12x−8−(x3+53)+6x2−11=0
=>12x−144=0
=>x=12
Vậy x=12x=12.
cho tôi đúng đi
bài 5:
1: \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{12x^3y^2:6xy^2}{18xy^5:6xy^2}=\dfrac{2x^2}{3y^3}\)
2: \(\dfrac{10xy-5x^2}{2x^2-8y^2}=\dfrac{5x\cdot2y-5x\cdot x}{2\left(x^2-4y^2\right)}\)
\(=\dfrac{5x\left(2y-x\right)}{-2\left(x+2y\right)\left(2y-x\right)}=\dfrac{-5x}{2\left(x+2y\right)}\)
3: \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)
\(=\dfrac{\left(x^2-xy\right)-\left(x-y\right)}{\left(x^2+xy\right)-\left(x+y\right)}\)
\(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
4: \(\dfrac{\left(x+1\right)\left(x^2-2x+1\right)}{\left(6x^2-6\right)\left(x^3-1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)^2}{6\left(x^2-1\right)\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)}{6\left(x-1\right)\left(x+1\right)\cdot\left(x^2+x+1\right)}\)
\(=\dfrac{1}{6\left(x^2+x+1\right)}\)
5: \(\dfrac{2x^2-7x+3}{1-4x^2}\)
\(=-\dfrac{2x^2-7x+3}{4x^2-1}\)
\(=-\dfrac{2x^2-6x-x+3}{\left(2x-1\right)\left(2x+1\right)}\)
\(=-\dfrac{2x\left(x-3\right)-\left(x-3\right)}{\left(2x-1\right)\left(2x+1\right)}\)
\(=-\dfrac{\left(x-3\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{-x+3}{2x+1}\)
Bài 3:
1: \(9x^3-xy^2\)
\(=x\cdot9x^2-x\cdot y^2\)
\(=x\left(9x^2-y^2\right)\)
\(=x\left(3x-y\right)\left(3x+y\right)\)
2: \(x^2-3xy-6x+18y\)
\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)
\(=x\left(x-3y\right)-6\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-6\right)\)
3: \(x^2-3xy-6x+18y\)
\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)
\(=x\left(x-3y\right)-6\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-6\right)\)
4: \(6xy-x^2+36-9y^2\)
\(=36-\left(x^2-6xy+9y^2\right)\)
\(=36-\left(x-3y\right)^2\)
\(=\left(6-x+3y\right)\left(6+x-3y\right)\)
5: \(x^4-6x^2+5\)
\(=x^4-x^2-5x^2+5\)
\(=x^2\left(x^2-1\right)-5\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x-1\right)\left(x+1\right)\)
6: \(9x^2-6x-y^2+2y\)
\(=\left(9x^2-y^2\right)-\left(6x-2y\right)\)
\(=\left(3x-y\right)\left(3x+y\right)-2\left(3x-y\right)\)
\(=\left(3x-y\right)\left(3x+y-2\right)\)