cho \(\dfrac{3a^2-b^2}{a^2+b^2}\) = \(\dfrac{3}{4}\). Tính \(\dfrac{a}{b}\)
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9 tháng 12 2022
2: \(\left(\dfrac{7}{a+7}+\dfrac{a^2+49}{a^2-49}-\dfrac{7}{a-7}\right):\dfrac{a+1}{2}\)
\(=\dfrac{7a-49+a^2+49-7a-49}{\left(a-7\right)\left(a+7\right)}\cdot\dfrac{2}{a+1}\)
\(=\dfrac{a^2-49}{\left(a-7\right)\left(a+7\right)}\cdot\dfrac{2}{a+1}=\dfrac{2}{a+1}\)
3: \(=\dfrac{x^4-4x^2+4x^2}{x^2-4}\cdot\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x\left(x^2-4\right)}\cdot\dfrac{x^2-4}{x-2}\right)\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x\left(x-2\right)}\right)\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x^2-4\right)+\left(2-3x\right)\left(x-4\right)}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-4x+2x-8-3x^2+12x}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-3x^2+10x-8}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-x^2-2x^2+2x+8x-8}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^3\left(x-1\right)\left(x^2-2x+8\right)}{\left(x-2\right)^2\cdot\left(x+2\right)\left(x-4\right)}\)
HN
24 tháng 4 2018
\(\dfrac{4}{a+b}-\dfrac{2a^2+3b^2}{2a^3+3b^3}-\dfrac{2b^2+3a^2}{2b^3+3a^3}=\dfrac{\left(a-b\right)^2.\left(12b^4+12ab^3-a^2b^2+12a^3b+12a^4\right)}{\left(a+b\right)\left(2a^3+3b^3\right)\left(2b^3+3a^3\right)}\ge0\)
PS: Còn cách dùng holder nữa mà lười quá
24 tháng 4 2018
holder Câu hỏi của Lê Minh Đức - Toán lớp 9 - Học toán với OnlineMath
DX
23 tháng 12 2022
2.
\(P=\left(\dfrac{a+6}{3\left(a+3\right)}-\dfrac{1}{a+3}\right).\dfrac{27a}{a+2}=\left(\dfrac{a+3}{3\left(a+3\right)}\right).\dfrac{27a}{a+2}=\dfrac{27a}{3\left(a+2\right)}=\dfrac{9a}{a+2}\)
ĐKXĐ là :
\(a\ne0;-3;-2\)
Vs a = 1 ta có:
=> P=3
1.
\(M=\left(\dfrac{2a}{2a+b}-\dfrac{4a^2}{\left(2a+b\right)^2}\right):\left(\dfrac{2a}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{1}{2a-b}\right)=\left(\dfrac{4a^2+2ab-4a^2}{\left(2a+b\right)^2}\right).\left(\dfrac{\left(2a+b\right)\left(2a-b\right)}{b}\right)=\dfrac{2a.\left(2a-b\right)}{\left(2a+b\right)}\)
H9
HT.Phong (9A5)
CTVHS
23 tháng 7 2023
a) \(\dfrac{3a^2}{10b^3}\cdot\dfrac{15b}{9a^4}\)
\(=\dfrac{3a^2\cdot15b}{10b^3\cdot9a^4}\)
\(=\dfrac{1\cdot3}{2\cdot b^2\cdot3\cdot a^2}=\dfrac{3}{6a^2b^2}\)
b) \(\dfrac{x-3}{x^2}\cdot\dfrac{4x}{x^2-9}\)
\(=\dfrac{x-3}{x^2}\cdot\dfrac{4x}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{\left(x-3\right)\cdot4x}{x^2\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{4}{x\left(x+3\right)}\)
c) \(\dfrac{a^2-6x+9}{a^2+3a}\cdot\dfrac{2a+6}{a-3}\)
\(=\dfrac{\left(a-3\right)^2}{a\left(a+3\right)}\cdot\dfrac{2\cdot\left(a+3\right)}{a-3}\)
\(=\dfrac{\left(a-3\right)^2\cdot2\cdot\left(a+3\right)}{a\left(a+3\right)\left(a-3\right)}\)
\(=\dfrac{2\left(a-3\right)}{a}\)
d) \(\dfrac{x+1}{x}\cdot\left(x+\dfrac{2-x^2}{x^2-1}\right)\)
\(=\dfrac{\left(x+1\right)\cdot x}{x}+\dfrac{x+1}{x}\cdot\dfrac{2-x^2}{x^2-1}\)
\(=x+1+\dfrac{x+1}{x}\cdot\dfrac{2-x^2}{\left(x+1\right)\left(x-1\right)}\)
\(=x+\dfrac{2-x^2}{x\left(x-1\right)}\)
22 tháng 8 2017
1)
\(\Leftrightarrow\left(x^2-2+\dfrac{1}{x^2}\right)+\left(y^2-2+\dfrac{1}{y^2}\right)+z^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2+z^2=0\)
\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\Rightarrow\left|x\right|=1\\y-\dfrac{1}{y}=0\Rightarrow\left|y\right|=1\\z=0\end{matrix}\right.\)
22 tháng 8 2017
dk\(x,y,z,a,b,c\ne0\)\(\left\{{}\begin{matrix}\dfrac{a}{x}=A\\\dfrac{b}{y}=B\\\dfrac{c}{z}=C\end{matrix}\right.\) \(\Rightarrow A,B,C\ne0\)
\(\left\{{}\begin{matrix}A+B+C=2\\\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}A^2+B^2+C^2+2\left(AB+BC+AC\right)=4\\\dfrac{ABC}{A}+\dfrac{ABC}{B}+\dfrac{ABC}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}AB+BC+AC=0\\A^2+B^2+C^2=4\end{matrix}\right.\)
\(\left(\dfrac{a}{x}\right)^2+\left(\dfrac{b}{y}\right)^2+\left(\dfrac{c}{z}\right)^2=4\)
PT
1
BK
6 tháng 12 2017
\(\dfrac{3a^2-b^2}{a^2+b^2}=\dfrac{3}{4}\)
\(\Rightarrow3\left(a^2+b^2\right)=4\left(3a^2-b^2\right)\)
\(\Rightarrow3a^2+3b^2=12a^2-4b^2\)
\(\Rightarrow-9a^2=-7b^2\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{-7}{-9}=\dfrac{7}{9}\Rightarrow\dfrac{a}{b}=\dfrac{\sqrt{7}}{3}\)
Vậy............
19 tháng 11 2018
a, \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)
\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}\)
19 tháng 11 2018
c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)
Ta có: \(\dfrac{1}{x+3}=\dfrac{1\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x^2-9}\)
\(\Rightarrow\dfrac{2}{x^2-9}+\dfrac{1}{x+3}=\dfrac{2}{x^2-9}+\dfrac{x-3}{x^2-9}=\dfrac{2+x-3}{x^2-9}=\dfrac{x-1}{x^2-9}\)
\(\Leftrightarrow12a^2-4b^2=3a^2+3b^2\)
\(\Leftrightarrow9a^2=7b^2\)
\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{7}{9}\)
hay \(\dfrac{a}{b}\in\left\{\dfrac{\sqrt{7}}{3};-\dfrac{\sqrt{7}}{3}\right\}\)
\(\dfrac{3a^2-b^2}{a^2+b^2}=\dfrac{3}{4}\)
\(\Leftrightarrow4.\left(3a^2-b^2\right)=3\left(a^2+b^2\right)\)
\(\Leftrightarrow12a^2-4b^2=3a^2+3b^2\)
\(\Leftrightarrow12a^2-3a^2=3b^2+4b^2\)
\(\Leftrightarrow9a^2=7b^2\)
\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{7}{9}\)
\(\text{hoặc }\dfrac{a}{b}=\pm\dfrac{\sqrt{7}}{3}\)