\(a,\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)

=>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. \(\dfrac{1}{3}.\left(x-1\right)+\dfrac{2}{5}.\left(x+1\right)=0\)

=> \(\dfrac{1}{3}x-\dfrac{1}{3}+\dfrac{2}{5}x+\dfrac{2}{5}=0\)

=> \(\dfrac{1}{3}x+\dfrac{2}{5}x=0+\dfrac{1}{3}-\dfrac{2}{5}\)

=> \(\dfrac{11}{15}x=\dfrac{-1}{15}\)

=> \(x=\dfrac{-1}{11}\)

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....

bai nay de\

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)

làm ơn giúp 🙏🙏🙏

a: =>1/3x+2/5x-2/5=0

=>11/15x-2/5=0

=>11/15x=2/5

=>x=2/5:11/15=2/5*15/11=30/55=6/11

b: =>-5x-1-1/2x+1/3=x

=>-11/2x-2/3-x=0

=>-13/2x=2/3

=>x=-2/3:13/2=-2/3*2/13=-4/39

c: (x+1/2)(2/3-2x)=0

=>x+1/2=0 hoặc 2/3-2x=0

=>x=1/3 hoặc x=-1/2

d: 9(3x+1)^2=16

=>(3x+1)^2=16/9

=>3x+1=4/3 hoặc 3x+1=-4/3

=>3x=1/3 hoặc 3x=-7/3

=>x=1/9 hoặc x=-7/9

a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3

=>x=-1/3+3/4=-4/12+9/12=5/12

b: =>x(1/2-5/6)=7/2

=>-1/3x=7/2

hay x=-21/2

c: (4-x)(3x+5)=0

=>4-x=0 hoặc 3x+5=0

=>x=4 hoặc x=-5/3

d: x/16=50/32

=>x/16=25/16

hay x=25

e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4

=>2x=-7/4+3=5/4

hay x=5/8

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

\(\Leftrightarrow\left(x-5\right)^2=2^2\)

\(\Leftrightarrow x-5=2\)

\(\Leftrightarrow x=2+5\)

\(\Leftrightarrow x=7\)

a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)

Th1 : \(x-\dfrac{1}{2}=0\)

         \(x=0+\dfrac{1}{2}\)

         \(x=\dfrac{1}{2}\)

Th2 : \(-3-\dfrac{x}{2}=0\)

         \(\dfrac{x}{2}=-3\)

         \(x=\left(-3\right)\cdot2\)

         \(x=-6\)

Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)

b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)

    \(x=\dfrac{5}{8}+\dfrac{1}{8}\)

   \(x=\dfrac{3}{4}\)

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

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

                \(\dfrac{3}{2}+x=\dfrac{3}{2}\)

                       \(x=\dfrac{3}{2}-\dfrac{3}{2}\)

                      \(x=0\)

d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)

    \(x+\dfrac{1}{3}=-4\)

    \(x=-4-\dfrac{1}{3}\)

    \(x=-\dfrac{13}{3}\)

ĐKXĐ: x>=0; x<>1

a: \(B=\dfrac{\sqrt{x}\left(x-1\right)^2}{\sqrt{x}+1}:\left(\left(x+\sqrt{x}+1+\sqrt{x}\right)\left(x-\sqrt{x}+1-\sqrt{x}\right)\right)\)

\(=\dfrac{\sqrt{x}\left(x-1\right)^2}{\sqrt{x}+1}:\left[\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2\right]\)

\(=\dfrac{\sqrt{x}\left(x-1\right)^2}{\left(x-1\right)^2\cdot\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)

b: Khi x=4-2căn 3=(căn 3-1)^2 thì \(B=\dfrac{\sqrt{3}-1}{\sqrt{3}-1+1}=\dfrac{\sqrt{3}-1}{\sqrt{3}}=\dfrac{3-\sqrt{3}}{3}\)

c: B=2/3

=>căn x/căn x+1=2/3

=>căn x=2

=>x=4

d: \(B-1=\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}=-\dfrac{1}{\sqrt{x}+1}< 0\)

=>B<1

e: B>1

=>-1/căn x+1>0

=>căn x+1<0(vô lý)

=>KO có x thỏa mãn

f: B nguyên khi căn x chia hết cho căn x+1

=>căn x+1-1 chia hết cho căn x+1

=>căn x+1=1 hoặc căn x+1=-1(loại)

=>căn x=0

=>x=0

giúp mik vs 4h 30 ) hc rồi