ta có:\(\frac{1}{2}a=\frac{2}{3}b=\frac{3}{4}c\)\(\Rightarrow\frac{1}{2}\times a\times\frac{1}{6}=\frac{2}{3}\times b\times\frac{1}{6}=\frac{3}{4}\times c\times\frac{1}{6}\)

\(\Rightarrow\frac{a}{12}=\frac{b}{9}=\frac{c}{8}=\frac{a-b}{12-9}=\frac{15}{3}=5\)

\(\Rightarrow\frac{a}{12}=5\Rightarrow a=12\times5=60\)

\(\Rightarrow\frac{b}{9}=5\Rightarrow b=9\times5=45\)

\(\Rightarrow\frac{c}{8}=5\Rightarrow c=8\times5=40\)

chúc bạn học tốt!!

\(\frac{1}{2}a=\frac{2}{3}b=\frac{3}{4}c=\frac{a}{2}=\frac{2b}{3}=\frac{3b}{4}\)

\(\Rightarrow\frac{a}{2.6}=\frac{2b}{3.6}=\frac{3c}{4.6}=\frac{a}{12}=\frac{b}{9}=\frac{c}{8}=\frac{a-b}{12-9}=\frac{15}{3}=5\)

\(\Rightarrow a=5.12=60\); \(b=5.9=45\); \(c=5.8=40\)

Vậy \(a=60\), \(b=45\), \(c=40\)

Bài 1:suy ra 5*(44-x)=3*(x-12)

                 220-5x=3x-36

                 -5x-3x=-36-220

                 -8x      =-256

                   x=32

Bài 2 :Đặt a/3=b/4=k

   suy ra a=3k ; b=4k

Ta có a*b=48

suy ra 3k*4k=48

         12k =48

         k=4

suy ra a=3*4=12

         b=4*4 =16 

Bài 3: áp dụng tính chất dãy số bằng nhau ta được 

    a+b+c+d/3+5+7+9 = 12/24=0,5

suy ra a=1,5;   b=2,5;    c=3,5;          d=4,

phiền quá đi

https://olm.vn/hoi-dap/detail/211794512831.html

Tham khảo ở link này (mình gửi cho)

Học tốt!!!!!!!!!!

a) \(\left(3+1\sqrt{6}-\sqrt{33}\right)\left(\sqrt{22}+\sqrt{6}+4\right)\)

\(=\sqrt{3}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right).\sqrt{2}\left(\sqrt{11}+\sqrt{3}+2\sqrt{2}\right)\)

\(=\sqrt{6}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right)\left(\sqrt{3}+2\sqrt{2}+\sqrt{11}\right)\)

\(=\sqrt{6}\left[\left(\sqrt{3}+2\sqrt{2}\right)^2-11\right]=\sqrt{6}\left(11+4\sqrt{6}-11\right)=\sqrt{6}.4\sqrt{6}=6.4=24\)

b) \(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)=\left(\frac{5+2\sqrt{6}+10-4\sqrt{6}}{5^2-\left(2\sqrt{6}\right)^2}\right)\left(15+2\sqrt{6}\right)\)

\(=\left(15-2\sqrt{6}\right)\left(15+2\sqrt{6}\right)=15^2-24=201\)

C) \(\left(\frac{4}{3}.\sqrt{3}+\sqrt{2}+\sqrt{3\frac{1}{3}}\right)\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{5}}\right)\)

\(=\left(\frac{4}{\sqrt{3}}+\frac{\sqrt{6}}{\sqrt{3}}+\frac{\sqrt{10}}{\sqrt{3}}\right)\left(\frac{\sqrt{6}}{\sqrt{5}}+\frac{\sqrt{10}}{\sqrt{5}}-\frac{4}{\sqrt{5}}\right)\)

\(=\frac{1}{\sqrt{15}}\left(\sqrt{6}+\sqrt{10}+4\right)\left(\sqrt{6}+\sqrt{10}-4\right)=\frac{1}{\sqrt{15}}\left[\left(\sqrt{6}+\sqrt{10}\right)^2-16\right]\)

\(=\frac{1}{\sqrt{15}}\left(16+4\sqrt{15}-16\right)=\frac{4\sqrt{15}}{\sqrt{15}}=4\)

d) \(\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1990+2\sqrt{1989}}=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1989+2\sqrt{1989}+1}\)

\(=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{\left(\sqrt{1989}+1\right)^2}=\left(\sqrt{1989}-1\right)\left(\sqrt{1989}+1\right)=1989-1=1988\)

e) \(\frac{a-\sqrt{ab}+b}{a\sqrt{a}+b\sqrt{b}}-\frac{1}{a-b}=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}-1}{a-b}\)

\(a)\)\(b^2-b+3\left(b+1\right)=0\)

\(\Leftrightarrow\)\(b^2-b+3b+3=0\)

\(\Leftrightarrow\)\(b^2+2b+1=-2\)

\(\Leftrightarrow\)\(\left(b+1\right)^2=-2\) ( vô lí vì \(\left(b+1\right)^2\ge0\) ) 

Vậy không có giá trị của b thỏa mãn đề bài 

Chúc bạn học tốt ~ 

\(b)\)\(\frac{4x-3}{2}=\frac{5-2x}{3}\)

\(\Leftrightarrow\)\(3\left(4x-3\right)=2\left(5-2x\right)\)

\(\Leftrightarrow\)\(12x-9=10-4x\)

\(\Leftrightarrow\)\(12x+4x=10+9\)

\(\Leftrightarrow\)\(16x=19\)

\(\Leftrightarrow\)\(x=\frac{19}{16}\)

Vậy \(x=\frac{19}{16}\)

Chúc bạn học tốt ~