\(a^2=bc\)

\(\Leftrightarrow\frac{a}{b}=\frac{c}{a}\)(\(b,a\ne0\))

\(\Leftrightarrow\frac{a}{b}-2=\frac{c}{a}-2\)

\(\Leftrightarrow\frac{a-2b}{b}=\frac{c-2a}{a}\)

áp dụng tính chất day tỉ số bằng nhau

thế như nào ý

Bài 1:

\(a)\)

\(B=-3xy^2.\frac{-2}{5}x^2y^3\)

\(=\frac{6}{5}.x^3y^5\)

Hệ số cao nhất: 1

Bậc của đơn thức: bậc 5

\(b)\)

Với: \(x=\left(-1\right);y=2\) ta được:

\(B=\frac{6}{5}\left(-1\right)^32^5=\frac{-192}{5}\)

Bài 2:

\(a)\)

\(A\left(x\right)=-3^2+5x+2x^4-8=2x^4-3x^2+5x-8\)

\(B\left(x\right)=-2x^4-8x+3x^2+3=-2x^4+3x^2-8x+3\)

\(b)\)

\(A\left(x\right)+B\left(x\right)=-3x-5\)

\(c)\)

\(A\left(x\right)-B\left(x\right)=4x^4-6x^2+13x-13\)

\(a)\)

\(\frac{2}{7}+\left(-\frac{3}{2}\right)+\left(-\frac{2}{5}\right)\)

\(=\frac{20}{70}+\frac{-105}{70}+\frac{-28}{70}\)

\(=\frac{-113}{70}\)

\(b)\)

\(-\frac{1}{3}+\frac{3}{5}+-\frac{1}{2}\)

\(=\frac{-10}{30}+\frac{18}{30}+\frac{-15}{30}\)

\(=\frac{-7}{30}\)

c, Đặt \(B\left(x\right)=\frac{x^3}{3}-5x=0\)

\(\Leftrightarrow x\left(\frac{x^2}{3}-5\right)=0\Leftrightarrow x\left(\frac{x^2-15}{3}\right)=0\)

\(\Leftrightarrow x\left(x^2-15\right)=0\Leftrightarrow x=\pm\sqrt{15};x=0\)

\(D=-\left|x+1\right|-\left|x-2\right|+6\)

Ta có:

\(\left|x+1\right|\ge0\Leftrightarrow-\left|x+1\right|\le0\)          (*)

\(\left|x-2\right|\ge0\Leftrightarrow-\left|x-2\right|\le0\)           (**)

Từ (*) và (**) ta được:

\(\Leftrightarrow-\left|x+1\right|-\left|x-2\right|\le0\)

\(\Leftrightarrow-\left|x+1\right|-\left|x-2\right|+6\le6\)

Vậy \(maxD=6\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

bạn có gửi mật khảu cho ai không

Ta có: \(a^2-1=a^2-a+a-1=\left(a+1\right)\left(a-1\right)\)

\(D=\frac{1}{2^2-1}+\frac{1}{3^2-1}+\frac{1}{4^2-1}+...+\frac{1}{20^2-1}\)

\(=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{19.21}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{2.4}+\frac{2}{3.5}+...+\frac{2}{19.21}\right)\)

\(=\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{4-2}{2.4}+\frac{5-3}{3.5}+...+\frac{21-19}{19.21}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\right)\)

\(=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{20}-\frac{1}{21}\right)\)

\(=\frac{589}{840}\)

\(E=\frac{1}{2^3-2}+\frac{1}{3^3-3}+\frac{1}{4^3-4}+...+\frac{1}{20^3-20}\)

\(=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{19.20.21}\)

\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{19.20.21}\right)\)

\(=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{21-19}{19.20.21}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{20.21}\right)\)

\(=\frac{209}{840}\)

40 độ nha

Vì đường thẳng aa' và bb' cắt nhau 1 điểm nên => $\widehat{aMb}$ đối đỉnh $\widehat{a^{\prime }M{b}^{\prime }}$ và $\widehat{bM{a}^{\prime }}$ đối đỉnh $\widehat{b^{\prime }Ma}$

=> Ta có : $\widehat{a^{\prime }Mx}=\widehat{xM{b}^{\prime }}=\frac{\widehat{a^{\prime }M{b}^{\prime }}}{2}=\frac{{80}^o}{2}$ = 40^o

Vậy $\widehat{a^{\prime }Mx}$ = 40^o và $\widehat{b^{\prime }Mx}$ = 40^o