\(A=\left(\dfrac{4x}{x+2}-\dfrac{x^3-8}{x^3+8}\times\dfrac{4x^2-8x+16}{x^2-4}\right)\div\dfrac{16}{x+2}\times\dfrac{x^2+3x+2}{x^2+x+1}\)
\(B=\dfrac{x^2+x-2}{x^3-1}\)
a) Tìm ĐKXĐ của A, B. Rút gọn A, B
b)Tìm GTLN của A+B
Tìm x biết
1) \(\left|\dfrac{1}{4}\times x^2+\dfrac{1}{45}\right|+\dfrac{1}{5}=\dfrac{1}{4}\)
2) \(\left(x^2-3\right)\times x^2-2\times x\times\left(x^2-3\right)\)
1: =>|1/4x^2+1/45|=1/20
=>1/4x^2+1/45=1/20 hoặc 1/4x^2+1/45=-1/20
=>1/4x^2=1/36
=>x^2=1/36:1/4=1/9
=>x=1/3 hoặc x=-1/3
2: =(x^2-3)(x^2-2x)
=x(x-2)(x^2-3)
Bài 1 Tìm \(x\)
a)\(\left(\dfrac{1}{2}\times x-3\right)\times\left(-\dfrac{1}{3}+x\right)=0\)
b)\(\dfrac{1}{2}\times x^2-\dfrac{1}{5}\times x=0\)
c)\(\dfrac{1`}{4}\times x=\dfrac{1}{16}\times x^2\)
d)\(9\times x^2=1\)
e)\(\left(x-5\right)^2=4\)
a) \(\left(\dfrac{1}{2}x-3\right)\left(-\dfrac{1}{3}+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=0\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=0+3\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{1}{2}\\x=0-\left(-\dfrac{1}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{1}{3}\end{matrix}\right.\)
d) \(9x^2=1\)
\(\Leftrightarrow x^2=1:9\)
\(\Leftrightarrow x^2=\dfrac{1}{9}\)
\(\Leftrightarrow x^2=\left(\dfrac{1}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
e) \(\left(x-5\right)^2=4\)
\(\Leftrightarrow\left(x-5\right)^2=2^2\)
\(\Leftrightarrow x-5=2\)
\(\Leftrightarrow x=2+5\)
\(\Leftrightarrow x=7\)
Tìm x
a, \(\left(2-x\right)\times\left(\dfrac{4}{5}-x\right)< 0\)
b, \(\left(x-\dfrac{3}{2}\right)\times\left(2\times x-1\right)>0\)
c, \(\left(x-\dfrac{4}{7}\right)\div\left(x+\dfrac{1}{2}\right)>0\)
d, \(\left(2\times x-5\right)\div\left(x+1\dfrac{3}{4}\right)< 0\)
Giúp mk với
\(a,\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)
=>Trong 2 số phải có 1 số âm và 1 số dương
Mà \(2-x>\dfrac{4}{5}-x\)
=>\(\dfrac{4}{5}< x< 2\)
Vậy...
a,\(\dfrac{1}{3}\times\left(x-1\right)+\dfrac{2}{5}\times\left(x+1\right)=0\)
b,4x-\(\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}x-5\right)\)
c,\(\left(x+\dfrac{1}{2}\right)\times\left(x-\dfrac{3}{4}\right)=0\)
Các bn ơi giúp mk với chiều mk đi học rồi !!!!!!!!!!
a. \(\dfrac{1}{3}.\left(x-1\right)+\dfrac{2}{5}.\left(x+1\right)=0\)
=> \(\dfrac{1}{3}x-\dfrac{1}{3}+\dfrac{2}{5}x+\dfrac{2}{5}=0\)
=> \(\dfrac{1}{3}x+\dfrac{2}{5}x=0+\dfrac{1}{3}-\dfrac{2}{5}\)
=> \(\dfrac{11}{15}x=\dfrac{-1}{15}\)
=> \(x=\dfrac{-1}{11}\)
Đây toán 8 mà? :v
a,\(\dfrac{1}{5}x\left(x-1\right)+\dfrac{2}{5}x\left(x+1\right)=0\)
\(\Leftrightarrow5x\left(x-1\right)+6x\left(x+1\right)=0\)
\(\Leftrightarrow\left[5\left(x-1\right)+6x\left(x+1\right)\right]x=0\)
\(\Leftrightarrow\left(5x-5+6x+6\right)x=0\)
\(\Leftrightarrow\left(11+1\right)x=0\)
\(\Leftrightarrow11x+1=0;x=0\)
\(\Leftrightarrow x=-\dfrac{1}{11};x=0\)
Vậy....
>; <; =?
a) \(\dfrac{2}{3}\times\dfrac{4}{5}\) 
\(\dfrac{4}{5}\times\dfrac{2}{3}\)
b) \(\left(\dfrac{1}{3}\times\dfrac{2}{5}\right)\times\dfrac{3}{4}\) 
\(\dfrac{1}{3}\times\left(\dfrac{2}{5}\times\dfrac{3}{4}\right)\)
c) \(\left(\dfrac{1}{3}+\dfrac{2}{15}\right)\times\dfrac{3}{4}\) 
\(\dfrac{1}{3}\times\dfrac{3}{4}+\dfrac{2}{15}\times\dfrac{3}{4}\)
a) \(\dfrac{2}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\dfrac{2}{3}\)
b) \(\left(\dfrac{1}{3}\times\dfrac{2}{5}\right)\times\dfrac{3}{4}=\dfrac{1}{3}\times\left(\dfrac{2}{5}\times\dfrac{3}{4}\right)\)
c) \(\left(\dfrac{1}{3}-\dfrac{2}{15}\right)\times\dfrac{3}{4}=\dfrac{1}{3}\times\dfrac{3}{4}+\dfrac{2}{15}\times\dfrac{3}{4}\)
thực hiện phép tính
a)\(\dfrac{2x^2-20x+50}{3x+3}\times\dfrac{x^2-1}{4\left(x-5\right)^2}\)
b) \(\dfrac{6x-3}{5x^2+x}\times\dfrac{25x^2+10x+1}{1-8x^3}\)
c) \(\dfrac{3x^2-x}{x^2-1}\times\dfrac{1-x^4}{\left(1-3x\right)^3}\)
a/ \(\dfrac{2x^2-20x+50}{3x+3}\cdot\dfrac{x^2-1}{4\left(x-5\right)^2}=\dfrac{2\left(x^2-10x+25\right)\cdot\left(x^2-1\right)}{3\left(x+1\right)\cdot4\left(x-5\right)^2}=\dfrac{2\left(x-5\right)^2\left(x-1\right)\left(x+1\right)}{12\left(x+1\right)\left(x-5\right)^2}=\dfrac{x+1}{6}\)
b/ \(\dfrac{6x-3}{5x^2+x}\cdot\dfrac{25x^2+10x+1}{1-8x^2}=-\dfrac{3\left(1-2x\right)\cdot\left(5x+1\right)^2}{x\left(5x+1\right)\left(1-2x\right)\left(1+2x+4x^2\right)}=\dfrac{3\left(5x+1\right)}{x\left(4x^2+2x+1\right)}\)
c/ \(\dfrac{3x^2-x}{x^2-1}\cdot\dfrac{1-x^4}{\left(1-3x\right)^3}=\dfrac{x-3x^2}{1-x^2}\cdot\dfrac{\left(1-x^2\right)\left(1+x^2\right)}{\left(1-3x\right)^3}=\dfrac{x\left(1-3x\right)\left(1-x^2\right)\left(1+x^2\right)}{\left(1-x^2\right)\left(1-3x\right)^3}=\dfrac{x\left(x^2+1\right)}{\left(1-3x\right)^3}\)
Dễ thế mà bạn ( người ko quen) ko làm đc !
Bài 1:
\(\dfrac{2}{1\times4}+\dfrac{2}{4\times7}+.....+\dfrac{2}{94\times97}\)
Bài 2:
\(6+3\times\left(x-3\right)=24\)
\(x\div0,25+\dfrac{x}{5}+x\times2=1972\)
a) \(\dfrac{-2}{15}\times x=\dfrac{-2}{7}\) b) \(\dfrac{7}{-5}\times x=-3\) c) \(-\dfrac{4}{9}x=\dfrac{1}{2}\) d) \(\dfrac{8}{3}\div x=\dfrac{-3}{8}\)
e) \(x\div\dfrac{3}{-4}=-12\) f) \(\left(-1\right)\div x=\dfrac{-3}{7}+\dfrac{4}{5}\) g)\(\dfrac{4}{11}x-\dfrac{1}{3}=\dfrac{2}{5}\) i) \(\dfrac{-6}{7}-\dfrac{1}{5}x=-4\)
j) \(\dfrac{1}{2}+\dfrac{2}{3}\div7=\dfrac{-1}{3}\) k) \(\dfrac{-5}{2}+x\div7=\dfrac{-1}{3}\) L) \(\dfrac{-3}{2}-\dfrac{1}{4}\div x=-1\)
a: =>x*2/15=2/7
=>x=2/7:2/15=2/7*15/2=15/7
b: x=3:7/5=15/7
c: x=-1/2:4/9=-1/2*9/4=-9/8
d: x=-8/3:3/8=-64/9
g: =>4/11x=2/5+1/3=6/15+5/15=11/15
=>x=11/15:4/11=121/60
l: =>1/4:x=1-3/2=-1/2
=>x=-1/4:1/2=-1/4*2=-1/2
k: =>x:7=-1/3+5/2=-2/6+15/6=13/6
=>x=91/6
