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Lời giải:
\(a+b=9\Rightarrow 2a+9=2a+(a+b)=3a+b\)
\(\Rightarrow \frac{6a+2b}{2a+9}=\frac{6a+2b}{3a+b}=\frac{2(3a+b)}{3a+b}=2(1)\)
\(a+b=9\Rightarrow b=9-a\Rightarrow -3a-b=-3a-(9-b)=-2a-9\)
\(\Rightarrow \frac{-3a-b}{-2a-9}=\frac{-2a-9}{-2a-9}=1(2)\)
Từ \((1);(2)\Rightarrow A=\frac{6a+2b}{2a+9}+\frac{-3a-b}{-2a-9}=2+1=3\)
Đặt ==k
Suy ra a=4k
b=9k
Ta có A=(3a -2b ≠ 0)
ð A=
A=
A==
Vậy A=
sorry sorry
đặt a/4=b/9=k
=> a=4k
b=9k
Ta có
A=4a-2b/3a-2b
A=4.4k-2.9k/3.4k-2.9k
A= k(16-18)/k(12-18)
A=-2/-6
A=1/3
Ta có:\(\frac{3a-b}{2a+15}=\frac{3a-b}{2a+a-b}=\frac{3a-b}{3a-b}=1\)
\(\frac{3b-a}{2b-15}=\frac{3b-a}{2b-\left(a-b\right)}=\frac{3b-a}{3b-a}=1\)
=>P=1+1=2
a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)
Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)
b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)
\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)
\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)
\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)
\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)
Do đó: +) \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)
+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)
+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)
\(\left(\sqrt{9}+\sqrt{4}\right)\sqrt{x}=10\)
\(\Rightarrow\left(3+2\right)\sqrt{x}=10\)
\(\Rightarrow5\cdot\sqrt{x}=10\) \(\Rightarrow\sqrt{x}=2\)
=> x = 4
Ta có: 2a = 2b = 2c => a = b = c
\(\Rightarrow A=\frac{a-b+c}{a+2b-c}=\frac{a-a+a}{a+2a-a}=\frac{a}{3a-a}=\frac{a}{2a}=\frac{1}{2}\)
1. \(\left(\sqrt{9}+\sqrt{4}\right)\sqrt{x}=10\)
\(\Rightarrow\left(3+2\right)\sqrt{x}=10\)
\(\Rightarrow5\sqrt{x}=10\)
\(\Rightarrow\sqrt{x}=2\)
\(\Rightarrow\left(\sqrt{x}\right)^2=2^2\)
\(\Rightarrow x=4\)
2. \(2a=2b=2c\)\(\Rightarrow a=b=c\)\(\Rightarrow A=\frac{a-b+c}{a+2b-c}=\frac{a-a+a}{a+2a-a}=\frac{a}{2a}=\frac{1}{2}\)
a) \(a>4\)cũng có thể là \(a\le-4\)
b) \(a\ge1\)cũng có thể là \(a\le-1\)
\(A=\frac{6a+2b}{2a+a+b}+\frac{3a+b}{2a+a+b}=\frac{9a+3b}{3a+b}=3\)