\(\left(-\frac{2}{3}+\frac{3}{7}\right):\frac{4}{5}+\left(-\frac{1}{3}+\frac{4}{7}\right):\frac{4}{5}\)

\(=\left(-\frac{2}{3}+\frac{3}{7}-\frac{1}{3}+\frac{4}{7}\right):\frac{4}{5}\)

\(=0:\frac{4}{5}\)

\(=0\)

ti ck nha

\(\left(\frac{-2}{3}+\frac{3}{7}\right):\frac{4}{5}+\left(\frac{-1}{3}+\frac{4}{7}\right):\frac{4}{5}\)

\(=\left(\frac{-2}{3}+\frac{3}{7}\right).\frac{5}{4}+\left(\frac{-1}{3}+\frac{4}{7}\right).\frac{5}{4}\)

\(=\frac{5}{4}.\left(\frac{-2}{3}+\frac{3}{7}+\frac{-1}{3}+\frac{4}{7}\right)\)

\(=\frac{5}{4}.\left(\frac{-3}{3}+\frac{7}{7}\right)\)

\(=\frac{5}{4}.\left(-1+1\right)\)

\(=\frac{5}{4}.0\)

\(=0\)

Vì x=-2 là no của f(x) => f(-2)=0 => m (-2)^2 +2.(-2)+16=0

=> 4m-4+16=0 => 4m=12 => m=3

f(x) = mx^2 + 2x + 16

=> f(-2) = m.(-2)^2 + 2.(-2) + 16 

=> f(-2) = 4m - 4 + 16

=> f(-2) = 4m + 12

=> 4m + 12 = 0

=> m = -3

Vậy tại f(-2)=mx^2+2x+16 thì m=-3

\(3\frac{1}{2}-4\frac{2}{3}\cdot\left(\frac{3}{4}-2\frac{1}{3}\right)-\left(\frac{5}{6}-\frac{7}{4}\right)+5\frac{1}{2}-3\)

\(=\frac{7}{2}-\frac{14}{3}\cdot\left(\frac{3}{4}-\frac{7}{3}\right)-\left(\frac{5}{6}-\frac{7}{4}\right)+\frac{11}{2}-3\)

\(=\frac{7}{2}-\frac{14}{3}\cdot\left(-\frac{19}{12}\right)+\frac{11}{12}+\frac{11}{2}-3\)

\(=\frac{7}{2}+\frac{133}{198}+\frac{77}{12}-3\)

\(=\frac{3005}{396}\)

Ta có

\(2a=3b=>\frac{a}{3}=\frac{b}{2}\)

áp dụng tính chất DTSBN ta có

\(\frac{a}{3}=\frac{b}{2}=\frac{a+b}{3+2}=\frac{15}{5}=3\)

\(+>\frac{a}{3}=3=>a=9\)

\(+>\frac{b}{2}=3=>b=6\)

ti ck nha

C1:Ta có 2a=3b=> a/b=3/2 => a=3k, b=2k

a+b=3k+2k=15 =>5k=15 => k=3

=> a=9, b=6

C2: Ta có 2a=3b=>a/3=b/2 =a+b/3+2=15/5= 3 (t/c dãy tỉ số bằng nhau)

=> a=9, b=6

C1: Đặt a/b=c/d=k => a=bk, c=dk

Ta có (a-b)^2/(c-d)^2=(bk-b)^2/(dk-d)^2 =b^2 . (k-2)^2 / d^2 .(k-1)^2 =b^2/d^2

         ab/cd=bk.b/dk.d =b^2/d^2

=> (a-b)^2/(c-d)^2=ab/cd

Dễ thấy 6,3 . 12 - 21 . 3,6 = 63 . 1,2 - 63 . 1,2 = 0

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