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\(\frac{2a}{a+b}+\frac{b}{a-b}=2< =>2\left(a-b\right)a+b\left(a+b\right)=2\left(a-b\right)\left(a+b\right).\)
\(< =>2a^2-2ab+ab+b^2=2a^2-2b^2\)
\(< =>3b^2-ab=0< =>b\left(3b-a\right)=0=>\orbr{\begin{cases}b=0\\3b-a=0\end{cases}}\)\(< =>\orbr{\begin{cases}b=0\\a=3b\end{cases}=>\orbr{\begin{cases}A=3\\A=1\end{cases}}}\)
`a/b=3/5=>a=3/5b`
Thay `a=3/5b` vào `[2a-5b]/[a-3b]` có:
`[2. 3/5b-5b]/[3/5b-3b]`
`=[6/5b-5b]/[3/5b-3b]`
`=[-19/5b]/[-12/5b]`
`=[-19/5]/[-12/5]=19/12`
\(\dfrac{2a-5b}{a-3b}=\dfrac{2\left(\dfrac{a}{b}\right)-5}{\left(\dfrac{a}{b}\right)-3}=\dfrac{2.\dfrac{3}{4}-5}{\dfrac{3}{4}-3}=\dfrac{14}{9}\)
\(\dfrac{a}{2}=\dfrac{b}{3}\Rightarrow b=\dfrac{3}{2}a\)
\(\dfrac{a}{2}=\dfrac{c}{5}\Rightarrow c=\dfrac{5}{2}a\)
=>B=\(\dfrac{a+7\cdot\left(\dfrac{3}{2}a\right)-2\cdot\left(\dfrac{5}{2}a\right)}{3a+2\cdot\left(\dfrac{3}{2}a\right)-\dfrac{5}{2}a}=\dfrac{a+\dfrac{21}{2}a-5a}{3a+3a-\dfrac{5}{2}a}=\dfrac{\dfrac{13}{2}a}{\dfrac{7}{2}a}=\dfrac{13}{7}\)
\(\dfrac{1}{2a-1}=\dfrac{2}{3b-1}=\dfrac{3}{4c-1}\Rightarrow\dfrac{2a-1}{1}=\dfrac{3b-1}{2}=\dfrac{4c-1}{3}\)
\(\Rightarrow\dfrac{36a-18}{18}=\dfrac{24b-8}{16}=\dfrac{12c-3}{9}\)và 3a+2b-c=4
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{36a-18}{18}=\dfrac{24b-8}{16}=\dfrac{12c-3}{9}=\dfrac{36a-18+24b-8-12c+3}{18+16-9}=\dfrac{12\left(3a+2b-c\right)-23}{25}=\dfrac{12\cdot4-23}{25}=1\)
=>2a-1=1<=>a=1
3b-1=2<=>b=1
4c-1=3<=>c=1
Vậy...
Mik làm luôn mik ko chép đề đâu
1)
a) \(\left|x\right|=\dfrac{5}{9}+\dfrac{3}{5}\)
\(\left|x\right|=\dfrac{25}{45}+\dfrac{27}{45}\)
\(\left|x\right|=\dfrac{52}{45}\)
\(\Rightarrow x=\dfrac{52}{45}or\left(-\dfrac{52}{45}\right)\)
Mà x>0
\(\Rightarrow x=\dfrac{52}{45}\)
Vậy \(x=\dfrac{52}{45}\)
b) \(-2\left|x\right|=\dfrac{-4}{3}\)
\(\left|x\right|=\dfrac{-4}{3}:\left(-2\right)\)
\(\left|x\right|=\dfrac{-4}{3}.\dfrac{-1}{2}\)
\(\left|x\right|=\dfrac{4}{6}\)
\(\left|x\right|=\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{2}{3}or\left(-\dfrac{2}{3}\right)\)
Mà x<0
\(\Rightarrow x=-\dfrac{2}{3}\)
Đặt 2a/5b=5b/6c=6c/7d=7d/2a=k
=> k^4=2a/5b.5b/6c.6c/7d.7d/2a=1
=>k=1 hoặc k=-1
Với k=1 thì B=4
Với k=-1 thì B=-4
Vậy B=4 hoặc B=-4
a) `A=a. 1/3 + a. 1/4 - a.1/6 = a. (1/3+1/4 -1/6)=a. 5/12`
Thay `a=-3/5: A=-3/5 . 5/12 =-1/4`
b) `B=b. 5/6+ b. 3/4-b. 1/2=b.(5/6+3/4-1/2)=b. 13/12`
Thay `b=12/13: B=12/13 . 13/12=1`.
a) Ta có: \(A=a\cdot\dfrac{1}{3}+a\cdot\dfrac{1}{4}-a\cdot\dfrac{1}{6}\)
\(=a\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{6}\right)\)
\(=a\cdot\left(\dfrac{4}{12}+\dfrac{3}{12}-\dfrac{2}{12}\right)\)
\(=a\cdot\dfrac{5}{12}\)
\(=\dfrac{-3}{5}\cdot\dfrac{5}{12}=\dfrac{-1}{4}\)
b) Ta có: \(B=b\cdot\dfrac{5}{6}+b\cdot\dfrac{3}{4}-b\cdot\dfrac{1}{2}\)
\(=b\left(\dfrac{5}{6}+\dfrac{3}{4}-\dfrac{1}{2}\right)\)
\(=b\cdot\left(\dfrac{10}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\)
\(=b\cdot\dfrac{5}{4}\)
\(=\dfrac{12}{13}\cdot\dfrac{5}{4}=\dfrac{60}{52}=\dfrac{15}{13}\)
Bài 1:
x/-3=9/4
nên x=-9/4*3=-27/4
2x+y=-4
=>y=-4-2x=-4-2*(-27/4)=-4+27/2=27/2-8/2=19/2
\(\dfrac{a}{b}=\dfrac{2}{3}\)=>3a=2b ; a=\(\dfrac{2}{3}b\)
=>\(\dfrac{3a+2b}{a+5b}=\dfrac{2b+2b}{\dfrac{2}{3}b+5b}=\dfrac{4b}{\dfrac{2}{3}b+\dfrac{15}{3}b}=\dfrac{4b}{\dfrac{17}{3}b}=\dfrac{12}{17}\)