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\(AB=\left(-6;-3\right)\)
\(CD=\left(x_D+2;-2\right)\)
Vì AB//CD
nên \(\dfrac{x_D+2}{-6}=\dfrac{-2}{-3}=\dfrac{2}{3}\)
\(\Leftrightarrow x_D+2=-4\)
hay \(x_D=-6\)
4/Giả xử \(\dfrac{a}{b}+\dfrac{b}{a}>=2\) (1)
<=> \(\dfrac{a^2+b^2}{ab}\)>=2
<=>a2+b2 >= 2ab
<=> a2+b2 - 2ab >=0
<=> (a-b)2 >= 0 (2)
Vì bđt (2) đúng nên bđt (1) đúng
b/ Gỉa sử \(\left(\dfrac{a+b}{2}\right)^2>=ab\)(1)
<=> \(\dfrac{\left(a+b\right)^2}{4}\)>= ab
<=> a2+b2+2ab>= 4ab
<=> a2+b2+2ab -4ab >=0
<=> (a-b)2>=0 (2)
Vidbđt (2) đúng nên bddt (1) đúng
c/Gỉa sử (ax+by)2<= (a2+b2)(x2+y2) (1)
<=> (ax)2+ (by)2+2*ax*by<= (ax)2 +(ay)2+(bx)2+(by)2
<=> 2*ax*by <= (ay)2+(bx)2
<=> 0<= (ay+bx)2(2)
(2) đúng nên(1) đúng
tui giúp đc nhiu đây thôi
1.
a. \(6x^4-9x^3=3x^3\left(2x-3\right)\)
b. \(x^2y^2z+xy^2z^2+x^2yz^2=xyz\left(xy+yz+xz\right)\)
d. \(2x\left(x+3\right)+2\left(x+3\right)=\left(x+3\right)\left(2x+2\right)=2\left(x+3\right)\left(x+1\right)\)
2b. \(4x\left(x+1\right)=8\left(x+1\right)\Leftrightarrow4x\left(x+1\right)-8\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(4x-8\right)=0\Leftrightarrow4\left(x+1\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy ...
2d. \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
Vậy ...
a)(4x-\(\dfrac{1}{2}\))3=(4x)3-3.(4x)2.\(\dfrac{1}{2}\)+3.4x.\(\left(\dfrac{1}{2}\right)\)2-(\(\dfrac12\))3=64x-24x2+3x-\(\dfrac18\)
b)(\(\dfrac{1}{2}\)x+5)3=(\(\dfrac{1}{2}x\))3+3.(\(\dfrac{1}{2}x\))2.5+3.\(\dfrac{1}{2}x\).52+53
=\(\dfrac18\)x+\(\dfrac{15}{4}\)x2+\(\dfrac{75}{2}x\)+125
c)8x3+\(\dfrac{1}{64}\)=(2x)3+(\(\dfrac14\))3=(2x+\(\dfrac14\))[(2x)2-2x.\(\dfrac{1}{4}\)+(\(\dfrac{1}{4}\))2 ]=(2x+\(\dfrac{1}{4}\))(4x2-\(\dfrac{1}{2}x\)+\(\dfrac{1}{16}\))