Sao lạ thế nhỉ, áp cái được luôn?

\(2a+\frac{b}{a}+\frac{c}{b}\ge3\sqrt[3]{2a.\frac{b}{a}.\frac{c}{b}}=3\sqrt[3]{2c}\)

Đẳng thức tự xét.

a.

\(\overrightarrow{u}=2\left(2;1\right)-\left(3;4\right)=\left(1;-2\right)\)

\(\overrightarrow{v}=3\left(3;4\right)-2\left(7;2\right)=\left(-5;8\right)\)

\(\overrightarrow{w}=5\left(7;2\right)+\left(2;1\right)=\left(37;11\right)\)

b.

\(\overrightarrow{x}=2\left(2;1\right)+\left(3;4\right)-\left(7;2\right)=\left(0;4\right)\)

\(\overrightarrow{z}=2\left(2;1\right)-3\left(3;4\right)+\left(7;2\right)=\left(2;-8\right)\)

c.

\(\overrightarrow{w}+\overrightarrow{a}=\overrightarrow{b}-\overrightarrow{c}\Rightarrow\overrightarrow{w}=\overrightarrow{b}-\overrightarrow{c}-\overrightarrow{a}\)

\(\Rightarrow\overrightarrow{w}=\left(3;4\right)-\left(7;2\right)-\left(2;1\right)=\left(-6;1\right)\)

AB//CD
=>\(\hat{A}+\hat{D}=180^0\)

mà \(2\cdot\hat{A}-\hat{D}=40^0\)

nên \(2\cdot\hat{A}-\hat{D}+\hat{A}+\hat{D}=180^0+40^0=220^0\)

=>\(3\cdot\hat{A}=220^0\)

=>\(\hat{A}=\left(\frac{220}{3}\right)^0\)

\(\hat{A}+\hat{D}=180^0\)

=>\(\hat{D}=180^0-\left(\frac{220}{3}\right)^0=\left(\frac{320}{3}\right)^0\)

AB//CD
=>\(\hat{B}+\hat{C}=180^0\)

mà \(\hat{B}-2\cdot\hat{C}=10^0\)

nên \(\hat{B}+\hat{C}-\hat{B}+2\cdot\hat{C}=180^0-10^0=170^0\)

=>\(3\cdot\hat{C}=170^0\)

=>\(\hat{C}=\left(\frac{170}{3}\right)^0\)

\(\hat{B}=180^0-\left(\frac{170}{3}\right)^0=\left(\frac{370}{3}\right)^0\)

1. A = ( -2a + 3b - 4c ) - ( -2a - 3b - 4c )

a) Rút gọn

A = ( -2a + 3b - 4c ) - ( -2a - 3b - 4c )

   = -2a + 3b - 4c + 2a + 3b + 4c

   = ( -2a + 2a ) + ( 3b + 3b ) + ( -4c + 4c )

   = 0 + 6b + 0

   = 6b

B1.a, A=6b

      b,A=-10851

B2.a,A=2b-2c

     b,B=2a

Đặt \(\frac{a}{b}=\frac{b}{c}=k\)

=>\(b=ck;a=bk=ck\cdot k=ck^2\)

\(\frac{a}{c}=\frac{ck^2}{c}=k^2\)

\(\frac{2a^2+5b^2}{2b^2+5c^2}=\frac{2\cdot\left(ck^2\right)^2+5\cdot\left(ck\right)^2}{2\cdot\left(ck\right)^2+5\cdot c^2}=\frac{c^2k^2\left(2k^2+5\right)}{c^2\left(2k^2+5\right)}=k^2\)

Do đó: \(\frac{a}{c}=\frac{2a^2+5b^2}{2b^2+5c^2}\) (ĐPCM)

x/y=y/z=z/x

=> x*z = 2*y = x*y = 2*z

Ta có :

x*z = x*y 

=> z=y

Ta có :

x*z = 2*y = y*y

Mà y = z (cmt)

=> x*z = y*z

=>x=y

Mà y = z (cmt)

=> x=y=z

\(\left(a+b\right).\left(b+c\right).\left(c-a\right)+\left(b+c\right).\left(c+a\right).\left(a-b\right)+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)

\(=\left(a+b\right).\left[\left(b+c\right).\left(c-a\right)+\left(c+a\right).\left(a-b\right)\right]+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)

\(=\left(a+b\right).\left(ac-a^2+bc-ab+a^2-ab+ac-bc\right)+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)

\(=-\left(a+b\right).2a.\left(b-c\right)+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)

\(=\left(a+b\right).\left(b-c\right).\left(-2a+c+a\right)=\left(a+b\right).\left(b-c\right).\left(c-a\right)\)

giai lai:

\(\left(b+c\right).\left[\left(a+b\right).\left(c-a\right)+\left(c+a\right).\left(a-b\right)\right]+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)

\(=-\left(b+c\right).2a.\left(b-c\right)+\left(b-c\right).\left(ac+bc+a^2+ab\right)\)

\(=\left(b-c\right).\left(-2ab-2ac+ac+bc+a^2+ab\right)\)

\(=\left(b-c\right).\left(-ab-ac+bc+a^2\right)\)

\(=\left(b-c\right).\left(a+b\right).\left(a-c\right)\)

\(a,A=\left(2a+b+3c\right)-\left(a-b+c\right)\)

\(=2a+b+3c-a+b-c\)

\(=a+2b-2c\)

\(b,B=\left(a+b-c\right)-\left(-2a+b-c\right)-\left(-a-b-2c\right)\)

\(=a+b-c+2a-b+c+a+b+2c\)

\(=4a+b+2c\)

\(c,C=\left(a-2b-c\right)-\left(-2a+b-c\right)-\left(-a-b-2c\right)\)

\(=a-2b-c+2a-b+c+a+b+2c\)

\(=4a-2b+2c\)

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