(X-2)²+(x-3)(x-2)=0
tìm x
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Thay t = 3 vào phương trình, ta được:
\(1-a-3=2a\left(a+2\right)\)
\(\Leftrightarrow-2-a=2a^2+4a\)
\(\Leftrightarrow2a^2+5a+2=0\)
Ta có \(\Delta=5^2-4.2.2=9,\sqrt{\Delta}=3\)
\(\Rightarrow\orbr{\begin{cases}a=\frac{-5+3}{4}=\frac{-1}{2}\\a=\frac{-5-3}{4}=-2\end{cases}}\)
\(2\left(x^4+y^4\right)\ge xy^3+x^3y+2x^2y^2\)
\(\Leftrightarrow\left(x^4-2x^2y^2+y^4\right)+\left(x^4-x^3y\right)+\left(y^4-xy^3\right)\ge0\)
\(\Leftrightarrow\left(x^2-y^2\right)^2+x^3\left(x-y\right)+y^3\left(y-x\right)\ge0\)
\(\Leftrightarrow\left(x^2-y^2\right)^2+\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)
\(\Leftrightarrow\left(x^2-y^2\right)^2+\left(x-y\right)^2\left[\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{2}\right]\ge0\) ( đúng )
\(\left(4x+3\right)^2=4\left(x-1\right)^2\)
\(\Leftrightarrow\left(4x+3\right)^2=\left(2x-2\right)^2\)
\(\Leftrightarrow16x^2+24x+9=4x^2-8x+4\)
\(\Leftrightarrow12x^2+32x+5=0\)
\(\Leftrightarrow12x^2+2x+30x+5=0\)
\(\Leftrightarrow2x\left(6x+1\right)+5\left(6x+1\right)=0\)
\(\Leftrightarrow\left(6x+1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}6x+1=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{6}\\x=\frac{-5}{2}\end{cases}}}\)
Vậy tập hợp nghiệm của pt \(S=\left\{\frac{-1}{6};\frac{-5}{2}\right\}\)
\(\left(4x+3\right)^2=4.\left(x-1\right)^2\)
\(\Leftrightarrow16x^2+24x+9=4.\left(x^2-2x+1\right)\)
\(\Leftrightarrow16x^2+24x+9=4x^2-8x+4\)
\(\Leftrightarrow16x^2+24x+9-4x^2+8x-4=0\)
\(\Leftrightarrow12x^2+32x+5=0\)
\(\Leftrightarrow12x^2+30x+2x+5=0\)
\(\Leftrightarrow6x\left(2x+5\right)+\left(2x+5\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(6x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+5=0\\6x+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-5}{2}\\x=\frac{-1}{6}\end{cases}}\)
Thay x = 4 vào phương trình, ta được :
\(1-m=2\left(2m+1\right)\left(m-1\right)\)
\(\Leftrightarrow2\left(2m+1\right)\left(m-1\right)+\left(m-1\right)=0\)
\(\Leftrightarrow\left(m-1\right)\left(4m+2+1\right)=0\)
\(\Leftrightarrow\left(m-1\right)\left(4m+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}m-1=0\\4m+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}m=1\\m=\frac{-3}{4}\end{cases}}\)
Hằng đẳng thức:\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Khi đó:
\(A=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{a^2+b^2+c^2-ab-bc-ca}\)
\(=a+b+c\)
\(=2011\)
1. 5( x^2 - 2x -1 ) + 2( 3x - 2) = 5( x + 10) ^2
<=> 5x^2 - 10x - 5 + 6x - 4 = 5 ( x^2 + 20x + 100)
<=> 5x^2 - 4x - 9 = 5x^2 + 100x + 500
<=> 5x^2 - 4x - 9 - 5x^2 - 100x - 500 = 0
<=> -104x - 509 = 0
<=> -104x = 509
<=> x = -509/104
Vậy S = { -509/104 }
\(\left(x-3\right)^3-2\left(x-1\right)=x\left(x-2\right)^2-5x^2\)
\(\Leftrightarrow x^3-9x^2+27x-27-2x+2=x\left(x^2-4x+4\right)-5x^2\)
\(\Leftrightarrow x^3-9x^2+25x-25=x^3-4x^2+4x-5x^2\)
\(\Leftrightarrow x^3-9x^2+25x-25=x^3-9x^2+4x\)
\(\Leftrightarrow21x-25=0\Leftrightarrow x=\frac{25}{21}\)
(=) (x-2)(x-2+x-3)=0
(=) (x-2)(2x-5)=0
(=) x-2=0 hoặc 2x-5=0
bn tự giải nha
#Học-tốt
\(\left(x-2\right)^2+\left(x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2+x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-5\right)=0\)
\(\Leftrightarrow x-2=0hoAC2x-5=0\)
\(\Leftrightarrow x\in\left\{2;\frac{5}{2}\right\}\)