Rut gon:
A=(x-2)^2-(2x+1)^2
B=(x-2y)^2-(x-2y) .(2y+x)
C=(x+1)^3-(x-2)^3
D=(x-1)^2-2(x-1)(x+1)+(x+1)^2
E=(x+2y)^2+2(x+2y)(x-2y)+(2y-x)
G=(2x+1)^3-(2x-1)
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D= (x-1) (x+5) (x2+4x+5)
=(x2+4x-5)(x2+4x+5)
=[x+4x]2-25\(\ge\)-25
Dấu "=" xảy ra khi x=0 hoặc x=-4
\(D=\left(x-1\right)\left(x+5\right)\left(x^2+4x+5\right)\)
\(D=\left(x^2+4x-5\right)\left(x^2+4x+5\right)\)
\(D=\left(x^2+4x\right)^2-5^2\)
\(D=\left(x^2+4x\right)^2-25\)
Vì \(\left(x^2-4x\right)^2\ge0\)
\(\Rightarrow\left(x^2-4x\right)^2-25\ge0-25\)
\(\Rightarrow D\ge-25\)
Vậy \(GTNN_D=-25\)tại \(x=0\)
Vậy
\(C=4x^2-4xy+y^2+4x^2-16x+16+1\)
\(=\left(2x-y\right)^2+(2x-4)^2+1\ge1\forall x;y\in R\)
Dấu "=" xảy ra<=> 2x-y=0 và 2x-4=0
<=>2x-y=0 và x=2 <=>y=4 và x=
Vậy....
\(B=3x^2-12x+16\)
\(=x^2-12x+36+2x^2-20\)
\(=\left(x-6\right)^2+2x^2-20\ge-20\forall x\in R\)
Dấu "=" xảy ra <=> \(\left(x-6\right)^2=0\)và \(2x^2=0\)
<=>x1 =6 và x2 =0
Vậy....
\(\frac{2}{x^2-4}-\frac{x-1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\)Đk \(x\ne\pm2;x\ne0\)
\(\Rightarrow\frac{2}{\left(x-2\right)\left(x+2\right)}-\frac{x-1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\)
\(\Rightarrow\frac{2x-\left(x-1\right)\left(x+2\right)+\left(x-4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=0\)
\(\Rightarrow2x-\left(x-1\right)\left(x+2\right)+\left(x-4\right)\left(x-2\right)=0\)
\(\Rightarrow2x-x^2-x+2+x^2-6x+8=0\)
\(\Rightarrow-5x+10=0\)
\(\Rightarrow-5x=-10\)
\(\Rightarrow x=2\)Loại
Ko có gt x thỏa mãn
\(\frac{1}{3-x}-\frac{1}{x+1}=\frac{x}{x-3}-\frac{\left(x-1\right)^2}{x^2-2x-3}\)
\(\Rightarrow\frac{1}{3-x}-\frac{1}{x+1}=\frac{x}{x-3}-\frac{\left(x-1\right)^2}{x^2-3x+x-3}\)
\(\Rightarrow\frac{1}{3-x}-\frac{1}{x+1}=\frac{x}{x-3}-\frac{\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}\)Đk \(x\ne3;x\ne-1\)
\(\Leftrightarrow\frac{1}{3-x}-\frac{1}{x+1}-\frac{x}{x-3}-\frac{\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}=0\)
\(\Rightarrow-\frac{1}{x-3}-\frac{1}{x+1}-\frac{x}{x-3}+\frac{\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}=0\)
\(\Rightarrow\frac{-1\left(x+1\right)-1\left(x-3\right)-x\left(x+1\right)+\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}=0\)
\(\Rightarrow-\left(x+1\right)-\left(x-3\right)-x\left(x+1\right)+\left(x-1\right)^2=0\)
\(\Rightarrow x-1-x+3-x^2-x+x^2-2x+1=0\)
\(\Rightarrow-3x+3=0\)
\(\Rightarrow-3x=-3\)
\(\Rightarrow x=1\)
\(4.\left(x-1\right)^2-9.\left(x+2\right)^2=0\)
\(\Rightarrow\left(2x-2\right)^2-\left(3x+6\right)^2=0\)
\(\Rightarrow\left(2x-2+3x+3\right).\left(2x-2-3x-6\right)=0\)
\(\Rightarrow\left(5x+1\right).\left(-x-8\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x+1=0\\-x-8=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{5}\\x=-8\end{cases}}\)
Vậy \(x=-\frac{1}{5}\text{ hoặc }x=-8\)
4.(x2-2.x.1+22) -9.(x2+2.x.2+22)=0 (4.x2-2.x.1.4+22.4)-(9.x2+2.x.2.9+22.4)=0 4.x2-8.x+16-9.x2-36.x -16=0 (4.x2-9.x2)-(8.x+36.x)+(16-16)=0 (-5.x2)-44.x+0=0 x.(-5x-44)=0-0 \(\hept{\begin{cases}x=0\\-5x-44=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\-5x=0+44\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\-5x=44\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\x=\frac{-44}{5}\end{cases}}\)
Ta có: \(T_1=2\pi\sqrt{\frac{m_1}{k}}\Rightarrow T_1^2=4\pi^2\frac{m_1}{k}\)
\(T_2=2\pi\sqrt{\frac{m_2}{k}}\Rightarrow T_2^2=4\pi^2\frac{m_2}{k}\)
Chu kì dao động của con lắc lò xo gồm cả 2 quả cầu là :
\(T=2\pi\sqrt{\frac{m_1+m_2}{k}}\Rightarrow T^2=4\pi^2[\frac{m_1}{k}+\frac{m_2}{k}]\)
Do đó :\(T=\sqrt{T_1^2+T_2^2}=\sqrt{0,6^2+0,8^2}=1,0\left(s\right)\)
Ta có :\(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
\(\Leftrightarrow a^2+b^2+c^2+3-2\left(a+b+c\right)\)
\(\Leftrightarrow a^2+b^2+c^2+3-2a-2b-2c=0\)
\(\Leftrightarrow a^2-2a+1+b^2-2b+1+c^2-2c+1=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
Vì bình phương luôn lớn hơn hoặc bằng 0
\(\Rightarrow\left(a-1\right)^2=0;\left(b-1\right)^2=0;\left(c-1\right)^2=0\)
\(\Leftrightarrow a=1;b=1;c=1\)
Vậy \(a=b=c=1\)
_Minh ngụy_
a2 + b2 + c2 + 3 = 2.( a + b + c )
\(\Leftrightarrow\) a2 + b2 + c2 + 3 - 2.( a + b + c ) = 0
\(\Leftrightarrow\) a2 + b2 + c2 + 3 - 2a - 2b - 2c = 0
\(\Leftrightarrow\) ( a2 - 2a + 1 ) + ( b2 - 2b + 1 ) + ( c2 - 2c + 1 ) = 0
\(\Leftrightarrow\) ( a - 1 )2 + ( b - 1 )2 + ( c - 1 )2 = 0
\(\Leftrightarrow\)\(\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}a-1=0\\b-1=0\\c-1=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}a=1\\b=1\\c=1\end{cases}}\)
Vậy a = 1 , b = 1 , c = 1
\(A=\left(x-2\right)^2-\left(2x+1\right)^2=x^2-4x+4-4x^2-4x-1=-3x^2+3=-3\left(x^2-1\right)\)
\(=-3\left(x-1\right)\left(x+1\right)\)
\(B=\left(x-2y\right)^2-\left(x-2y\right)\left(x+2y\right)=\left(x-2y\right)\left(x-2y-x-2y\right)=-4y\left(x-2y\right)\)
\(C=\left(x+1\right)^3-\left(x-2\right)^3=\left(x^3+3x^2+3x+1\right)-\left(x^3-6x^2+12x-8\right)\)
\(=x^3+3x^2+3x+1-x^3+6x^2-12x+8=9x^2-9x+9=9\left(x^2-x+1\right)\)
\(D=\left(x-1\right)^2-2\left(x-1\right)\left(x+1\right)+\left(x+1\right)^2=\left(x-1-x-1\right)^2=-2^2=4\)
\(E=\left(x+2y\right)^2+2\left(x+2y\right)\left(x-2y\right)+2y-x=x^2+4xy+4y^2+2\left(x^2-4y^2\right)+2y-x\)
\(=x^2+4xy+4y^2+2x^2-8y^2+2y-x=3x^2-4y^2+4xy+2y-x\)
\(G=\left(2x+1\right)^3-\left(2x-1\right)=8x^3+12x^2+6x+1-2x+1=8x^3+12x^2+4x+2\)
\(=2\left(4x^3+6x^2+2x+1\right)=2\left(4x\left(x+1\right)^2+1\right)\)