a/\(\left(\dfrac{7}{9}\times\dfrac{9}{7}\right)\times\dfrac{25}{28}\)
\(=1\times\dfrac{25}{28}\)
\(=\dfrac{25}{28}\)
b/\(\dfrac{4}{7}\times\dfrac{17}{18}\times\dfrac{7}{4}\times\dfrac{18}{17}\)
\(=\left(\dfrac{4}{7}\times\dfrac{7}{4}\right)\times\left(\dfrac{17}{18}\times\dfrac{18}{17}\right)\)
\(=1\times1\)
\(=1\)

lỗi

x + 4/9 + 8/18 + 12/27 + 16/36 = 16/9

 x + (4/9+12/27)+(8/12+16/36) = 16/9

 x +       8/9         +     10/9        = 16/9

  x +               2                         = 16/9

  x                                              = 16/9 - 2

  x                                               =  -2/18  

x+4/9+4/9+4/9+4/9=16/9
x+4/9.4=16/9
x+16/9=16/9
-->    x = 0

a. \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\cdot\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)

<=> \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

<=> \(P=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

<=> \(P=\dfrac{\sqrt{x}+2}{x-2\sqrt{x}}\)

b. Khi \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\) => \(\sqrt{x}=2+\sqrt{3}\)

=> \(P=\dfrac{2+\sqrt{3}+2}{7+4\sqrt{3}-2\left(2+\sqrt{3}\right)}=\dfrac{4+\sqrt{3}}{7+4\sqrt{3}-4-2\sqrt{3}}=\dfrac{4+\sqrt{3}}{3+2\sqrt{3}}=\dfrac{5\sqrt{3}-6}{3}\)

check giùm mik

a: Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\cdot\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)

\(=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2+4}{x-4}\)

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

b: Thay \(x=7+4\sqrt{3}\) vào P, ta được:

\(P=\dfrac{2+\sqrt{3}+2}{\sqrt{3}\left(2+\sqrt{3}\right)}=\dfrac{-6+5\sqrt{3}}{3}\)

ahihi

Cái này dễ lắm. Mình giải luôn nhé!

a) \(\left[{}\begin{matrix}\dfrac{1}{7}x-\dfrac{2}{7}=0\Leftrightarrow x=\dfrac{2}{7}:\dfrac{1}{7}\Leftrightarrow x=2\\-\dfrac{1}{5}x+\dfrac{3}{5}=0\Leftrightarrow x=-\dfrac{3}{5}:\left(-\dfrac{1}{5}\right)\Leftrightarrow x=3\\\dfrac{1}{3}x+\dfrac{4}{3}=0\Leftrightarrow x=-\dfrac{4}{3}:\dfrac{1}{3}\Leftrightarrow x=-4\end{matrix}\right.\)

Vậy x=2 hoặc x=3 hoặc x=-4

b)\(x\left(\dfrac{1}{6}+\dfrac{1}{10}-\dfrac{4}{15}\right)+1=0\)

\(x.0+1=0\)

\(1=0\) ( vô lí)

Vậy không có giá trị của x nào thỏa mãn

=>x-1/2=3+1/14-5-1/3-4/7+11/21

=>x-1/2=-97/42

=>x-21/42=-97/42

=>x=-97/42+21/42=-76/42=-38/21

\(a,\frac{7}{x}=\frac{x}{28}=>x\cdot x=28\cdot7=>x^2=196=>x^2=14^2\)\(=>x=14\)

\(b,\frac{10+x}{x+17}=\frac{3}{4}=>\left(10+x\right)\cdot4=\left(x+17\right)\cdot3=>40+x4=x3+51\)\(=>x4-x3=51-40=>x=11\)

a)TH1 x>=3 \(\left|x-3\right|\)=x-3

pttt: x-3-2x=1 suy ra x=-4 <3 -> loại

TH2 x=< 3 pttt 3-x-2x=1 suy ra x =2/3 thỏa mãn

b) VT=\(\dfrac{4^{x+2}+4^{x+1}+4^x}{21}=\dfrac{4^x\left(4^2+4+1\right)}{21}=4^x\)

VP= \(\dfrac{3^{2x}+3^{2x+1}+3^{2x+3}}{31}=\dfrac{9^x\left(1+3+27\right)}{31}=9^x\)

vậy pt đã cho tương đương với 4^x=9^x ^{\(\Leftrightarrow\left(\dfrac{4}{9}\right)\)}^x =1 suy ra x =0

các bn lm đến đâu cx dc miễn là lm hộ mk cái ạ, ai đang lm vào nhắn tin vs mk để mk bít nha

a; \(-\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{71}{15}< x< -\dfrac{13}{7}+\dfrac{19}{14}-\dfrac{7}{2}\)

              -\(\dfrac{19}{15}\) - \(\dfrac{71}{15}\) < \(x\) < -\(\dfrac{1}{2}\) - \(\dfrac{7}{2}\)

              -6 < \(x\) < -4

             vì \(x\) \(\in\) Z nên \(x\) = -5

- \(\dfrac{19}{6}+\dfrac{-15}{2}+\dfrac{11}{3}\) < \(x\) < - \(\dfrac{5}{4}\) + \(\dfrac{19}{12}\) - \(\dfrac{10}{3}\)

- \(\dfrac{32}{3}+\dfrac{11}{3}< x< \) \(\dfrac{1}{3}\) - \(\dfrac{10}{3}\)

-7 < \(x\) < - 3

Vì \(x\in\) Z nên \(x\in\) {-6;-5;-4}

a)

\(\frac{864\cdot48-432\cdot96}{864\cdot48\cdot432}=\frac{432\cdot2\cdot48-432\cdot96}{864\cdot48\cdot432}=\frac{432\cdot96-432\cdot96}{864\cdot48\cdot432}=\frac{0}{864\cdot48\cdot432}=0\)

Làm từng phần một nha =))

b)

\(\frac{16\cdot17-5}{16\cdot16+11}=\frac{16\cdot\left(16+1\right)-5}{16\cdot16+11}=\frac{16^2+16-5}{16^2+11}=\frac{16^2+11}{16^2+11}=1\)

@_@

câu khó quá mmk ko làm đc