bn đăng bên toán nhé

câu c) mang tính mua vui hay gì hả bn

mếu thật thì x=0,x=số nào cx đc(câu trả lời này mang tính mua vui thôi nhé)

mình cần gấp mong các bạn giải giùm

c: \(E=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)

`a)1/7xx2/7+1/7xx5/7+6/7`

`=1/7xx(2/7+5/7)+6/7`

`=1/7xx1+6/7`

`=1/7+6/7=1`

`b)6/11xx4/9+6/11xx7/9-6/11xx2/9`

`=6/11xx(4/9+7/9-2/9)`

`=6/11xx9/9`

`=6/11`

Sorry nãy ghi thiếu.

`c)4/25xx5/8xx25/4xx24`

`=(4xx5xx25xx24)/(25xx8xx4)`

`=(4xx5xx24)/(4xx8)`

`=(5xx24)/8`

`=5xx3=15`

a, \(\dfrac{1}{7}.\dfrac{2}{7}+\dfrac{1}{7}.\dfrac{5}{7}+\dfrac{6}{7}\)

\(=\dfrac{1}{7}.\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\dfrac{6}{7}\)
\(=\dfrac{1}{7}.1+\dfrac{6}{7}\)

\(=\dfrac{1}{7}+\dfrac{6}{7}=1\)

b, \(\dfrac{6}{11}.\dfrac{4}{9}+\dfrac{6}{11}.\dfrac{7}{9}-\dfrac{6}{11}.\dfrac{2}{9}\)

\(=\dfrac{6}{11}.\left(\dfrac{4}{9}+\dfrac{7}{9}-\dfrac{2}{9}\right)\)

\(=\dfrac{6}{11}.1=\dfrac{6}{11}\)

c, \(\dfrac{4}{25}.\dfrac{5}{8}.\dfrac{25}{4}.24\)

\(=\left(\dfrac{4}{25}.\dfrac{25}{4}\right).\left(\dfrac{5}{8}.24\right)\)

\(=1.15=15\)

\(a,5x\dfrac{7}{3}=\dfrac{5}{1}x\dfrac{7}{3}=\dfrac{35}{3};b,\dfrac{13}{4}:7=\dfrac{13}{4} :\dfrac{7}{1}=\dfrac{13}{4}x\dfrac{1}{7}=\dfrac{13}{28}\)

\(\dfrac{3}{7}+\dfrac{2}{5}+\dfrac{3}{4}=\dfrac{60}{140}+\dfrac{56}{140}+\dfrac{105}{140}=\dfrac{221}{140}\)

\(\dfrac{9}{7}-\dfrac{5}{11}x\dfrac{11}{7}=\dfrac{9}{7}-\dfrac{5}{7}=\dfrac{4}{7}\)

\(\dfrac{4-x}{-5}=\dfrac{-5}{4-x}\)

\(\left(4-x\right)^2=25=5^2=\left(-5\right)^2\)

4-x=5 hoặc 4-x=-5

x=-1 hoặc x=9

a) \(x=0,05\)

b) \(x=1,125\)

c) \(x=0,96\)

d) \(x=0,025\)

Bạn tự làm đi dễ mà . Cố mag vận động đầu óc đừng copy làm bài nữa khó lắm mới hỏi thôi

Chuẩn đấy

1: \(P=\left(\dfrac{2x}{x^2-9}-\dfrac{1}{x+3}\right):\left(\dfrac{2}{x}-\dfrac{x-1}{x^2-3x}\right)\)

\(=\left(\dfrac{2x}{\left(x-3\right)\left(x+3\right)}-\dfrac{1}{x+3}\right):\left(\dfrac{2}{x}-\dfrac{x-1}{x\cdot\left(x-3\right)}\right)\)

\(=\dfrac{2x-x+3}{\left(x-3\right)\left(x+3\right)}:\dfrac{2\left(x-3\right)-x+1}{x\left(x-3\right)}\)

\(=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x\left(x-3\right)}{2x-6-x+1}\)

\(=\dfrac{x}{x-5}\)

\(\dfrac{x-2}{x+1}-\dfrac{3}{x+2}>0.\left(x\ne-1;-2\right).\\ \Leftrightarrow\dfrac{x^2-4-3x-3}{\left(x+1\right)\left(x+2\right)}>0.\\ \Leftrightarrow\dfrac{x^2-3x-7}{\left(x+1\right)\left(x+2\right)}>0.\)    

Đặt \(f\left(x\right)=\dfrac{x^2-3x-7}{\left(x+1\right)\left(x+2\right)}>0.\)

Ta có: \(x^2-3x-7=0.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{37}}{2}.\\x=\dfrac{3-\sqrt{37}}{2}.\end{matrix}\right.\)

          \(x+1=0.\Leftrightarrow x=-1.\\ x+2=0.\Leftrightarrow x=-2.\)

Bảng xét dấu:

\(\Rightarrow f\left(x\right)>0\Leftrightarrow x\in\left(-\infty-2\right)\cup\left(\dfrac{3-\sqrt{37}}{2};-1\right)\cup\left(\dfrac{3+\sqrt{37}}{2};+\infty\right).\)

\(\sqrt{x^2-3x+2}\ge3.\\ \Leftrightarrow x^2-3x+2\ge9.\\ \Leftrightarrow x^2-3x-7\ge0.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3-\sqrt{37}}{2}.\\x=\dfrac{3+\sqrt{37}}{2}.\end{matrix}\right.\)

Đặt \(f\left(x\right)=x^2-3x-7.\)

\(f\left(x\right)=x^2-3x-7.\)

\(\Rightarrow f\left(x\right)\ge0\Leftrightarrow x\in(-\infty;\dfrac{3-\sqrt{37}}{2}]\cup[\dfrac{3+\sqrt{37}}{2};+\infty).\)

\(\Rightarrow\sqrt{x^2-3x+2}\ge3\Leftrightarrow x\in(-\infty;\dfrac{3-\sqrt{37}}{2}]\cup[\dfrac{3+\sqrt{37}}{2};+\infty).\)