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\(\left|x-1\right|+\left|2x-2\right|+\left|3x-3\right|=6\left(1\right)\)
Xét : \(x-1=0\Leftrightarrow x=1;x-1< 0\Leftrightarrow x< 1;x-1>0\Leftrightarrow x>1\)
\(2x-2=0\Leftrightarrow x=1;2x-2< 0\Leftrightarrow x< 1;2x-2>0\Leftrightarrow x>1\)
\(3x-3=0\Leftrightarrow x=1;3x-3< 0\Leftrightarrow x< 1;3x-3>0\Leftrightarrow x>1\)
Ta có bảng xét dấu các đa thức x-1 ; 2x-2 ; 3x-3 sau :
X | 1 |
x-1 | - 0 + |
2x-2 | - 0 + |
3x-3 | - 0 + |
Xét khoảng \(x< 1\) ta có :
(1) \(\Leftrightarrow1-x+2-2x+3-3x=6\Leftrightarrow6-6x=6\Leftrightarrow x=0\) (Giá trị này thuộc khoảng đang xét )
Xét khoảng \(x>0\) ta có :
(1) \(\Leftrightarrow x-1+2x-2+3x-3=6\Leftrightarrow6x-6=6\Leftrightarrow x=2\) ( Giá trị này thuộc khoảng đang xét )
Vậy \(x=0\) và \(x=2\) thỏa mãn
\(a,\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)=9x^2+30x+25+9x^2-30x+25-9x^2+4=9x^2+54\)
\(b,BT=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x=x^3-16x^2+25x\)
\(c,BT=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-z-x-y\right)^2=z^2\)
\(\left(3x-1\right)^2+2\left(9x^2-1\right)+\left(3x+1\right)^2\)
\(=9x^2-6x+1+18x^2+2+9x^2+6x+1\)
\(=36x^2+4\)
\(\left(x^2-1\right)\left(x+3\right)-\left(x-3\right)\left(x^3+3x+9\right)\)
\(=x^3+3x^2-x+3-\left(x^4+3x^2+9x-3x^3-9x-27\right)\)
\(=x^3+3x^2-x+3-x^4-3x^2-9x+3x^3+9x-27\)
\(=\left(3x^2-3x^2\right)+\left(9x-9x\right)-x-\left(27-3\right)+x^3-x^4+3x^3\)
\(=-x-24+x^3-x^4+3x^3\)
\(\left(x+4\right)\left(x-4\right)-\left(x-4\right)^2\)
\(=x^2-16-\left(x-4\right)^2\)
\(=x^2-16-x^2+8x-16\)
\(=8x-32\)
\(\Leftrightarrow\dfrac{1}{2}\left(x^2-4x+4\right)-\dfrac{13}{3}\left(x^2+6x+9\right)=\dfrac{1}{4}\left(x^2-3x+2\right)-2\left(9x^2+3x-2\right)\)
\(\Leftrightarrow x^2\cdot\dfrac{1}{2}-2x+2-\dfrac{13}{3}x^2-26x-39=\dfrac{1}{4}x^2-\dfrac{3}{4}x+\dfrac{1}{2}-18x^2-6x+4\)
\(\Leftrightarrow x^2\cdot\dfrac{167}{12}-\dfrac{85}{4}x-\dfrac{83}{2}=0\)
\(\Leftrightarrow167x^2-255x-498=0\)
\(\text{Δ}=\left(-255\right)^2-4\cdot167\cdot\left(-498\right)=397689\)
Vì Δ>0 nên phương trình có 2 nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{255-\sqrt{397689}}{334}\\x_2=\dfrac{255+\sqrt{397689}}{334}\end{matrix}\right.\)
\(H\left(x\right)=P\left(x\right)+Q\left(x\right)-R\left(x\right)\)
\(H\left(x\right)=\left(x^2+6x-2\right)+\left(3x^2-x+7\right)-\left(3x^2+x-1\right)\)
\(H\left(x\right)=x^2+6x-2+3x^2-x+7-3x^2-x+1\)
\(H\left(x\right)=\left(x^2-3x^2+3x^2\right)+\left(6x-x-x\right)+\left(-2+1+1\right)\)
\(H\left(x\right)=x^2+4x\)
Tìm nghiệm:
\(H\left(x\right)=x^2+4x\)
\(\Leftrightarrow x^2+4x=0\)
\(\Leftrightarrow x\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
a)
\(A=\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)
\(=x^3-3x^2+9x+3x^2-9x+27-54-x^3\)
\(=-27\)
or
\(A=x^3+27-54-x^3=-27\)
b)
\(B=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3=2y^3\)
c)
\(C=\left(2x+1\right)^2+\left(1-3x\right)^2+2\left(2x+1\right)\left(3x-1\right)\)
\(=\left(2x+1+3x-1\right)^2=\left(5x\right)^2=25x^2\)
d)
\(D=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=x^3-8-\left(x-1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=6x^2-3x-10\)
Lời giải:
a)
\((4x-1)^2+(3x+1)^2+2(4x-1)(3x+1)\)
\(=(4x-1)^2+2(4x-1)(3x+1)+(3x+1)^2\)
\(=[(4x-1)+(3x+1)]^2=(7x)^2=49x^2\)
b)
\((x^2+2)(x-5)+(x-5)(x^2+5x+25)\)
\(=(x-5)[(x^2+2)+(x^2+5x+25)]\)
\(=(x-5)(2x^2+5x+27)\)
Ta có:
\(x^2=\left(x-1\right)\left(3x-2\right)\)
\(\Rightarrow x^2=3x^2-5x+2\)
\(\Rightarrow3x^2-5x+2-x^2=0\)
\(\Rightarrow2x^2-4x-x+2=0\)
\(\Rightarrow\left(2x^2-4x\right)-\left(x-2\right)=0\)
\(\Rightarrow2x\left(x-2\right)-\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(2x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\2x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\2x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}\)
Vậy x = 2 hoặc x = 1/2
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