a: \(\Leftrightarrow\left(2x-2\right)\cdot\dfrac{3}{4}=-6-\dfrac{1}{3}=-\dfrac{19}{3}\)

\(\Leftrightarrow2x-2=-\dfrac{19}{3}:\dfrac{3}{4}=-\dfrac{76}{9}\)

=>2x=-58/9

hay x=-29/9

b: \(\Leftrightarrow\left(\dfrac{44}{7}x+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)

\(\Leftrightarrow x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{5}{7}\)

\(\Leftrightarrow x\cdot\dfrac{44}{7}=-\dfrac{8}{7}\)

hay x=-2/11

e: \(\left(2x+\dfrac{3}{5}\right)^2=\dfrac{9}{25}\)

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

|2x-1|=x+3

=> 2x-1=x+3 hoặc 2x-1=-(x+3)

2x-x=1+4 2x-1=-x-3

x=5 2x+x= 1-3

3x=-2

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

|4x+7|=2x+5

=> 4x+7=2x+5

4x-2x=5-7

-2x=-2

x=1

=>4x+7=-(2x+5)

4x+7=-2x-5

4x+2x=-5-7

6x=-12

x=-2

a: 1-2x<7

=>-2x<6

hay x>-3

b: (x-1)(x-2)>0

=>x-2>0 hoặc x-1<0

=>x>2 hoặc x<1

c: \(\left(x-2\right)^2\cdot\left(x+1\right)\left(x-4\right)< 0\)

=>(x+1)(x-4)<0

=>-1<x<4

a) \(2x\left(x-\frac{1}{7}\right)=0\)

⇒\(\left[{}\begin{matrix}2x=0\\x-\frac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{7}\end{matrix}\right.\)

Vậy \(x=0;x=\frac{1}{7}\)

b) \(\frac{1}{2}x+\frac{3}{5}x=\frac{-33}{25}\\ \left(\frac{1}{2}+\frac{3}{5}\right)x=\frac{-33}{25}\\ \left(\frac{5}{10}+\frac{6}{10}\right)x=\frac{-33}{25}\\ \frac{11}{10}x=\frac{-33}{25}\\ x=\frac{-33}{25}:\frac{11}{10}\\ x=\frac{-33.10}{25.11}\\ x=\frac{-6}{5}\)

Vậy x = \(\frac{-6}{5}\)

c) \(\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}+\frac{-3}{7}:x\right)=0\\ \Rightarrow\left[{}\begin{matrix}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}+\frac{-3}{7}:x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x=\frac{4}{9}\\\frac{-3}{7}:x=\frac{-1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{4}{9}:\frac{2}{3}=\frac{4.3}{9.2}=\frac{2}{3}\\x=\frac{-3}{7}:\frac{-1}{2}=\frac{-3.2}{7.\left(-1\right)}=\frac{6}{7}\end{matrix}\right.\)

a) \(2x\left(x-\frac{1}{7}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-\frac{1}{7}=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0:2=0\\x=0+\frac{1}{7}=\frac{1}{7}\end{matrix}\right.\)

b) \(\frac{1}{2}x+\frac{3}{5}x=-\frac{33}{25}\)

\(\Rightarrow x\left(\frac{1}{2}+\frac{3}{5}\right)=-\frac{33}{25}\)

\(\Rightarrow x\frac{11}{10}=-\frac{33}{25}\)

\(\Rightarrow x=\left(-\frac{33}{25}\right):\frac{11}{10}=-\frac{33}{25}.\frac{10}{11}=-\frac{6}{5}\)

c) \(\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}+\frac{-3}{7}:x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}+\frac{-3}{7}:x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x=0+\frac{4}{9}=\frac{4}{9}\\-\frac{3}{7}:x=0-\frac{1}{2}=-\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{4}{9}:\frac{2}{3}=\frac{4}{9}.\frac{3}{2}=\frac{2}{3}\\x=\left(-\frac{3}{7}\right):\frac{-1}{2}=\left(-\frac{3}{7}\right).\left(-2\right)=\frac{6}{7}\end{matrix}\right.\)

a: \(x^2+4x-1=0\)

=>\(x^2+4x+4-5=0\)

=>\(\left(x+2\right)^2=5\)

=>\(\left[\begin{array}{l}x+2=\sqrt5\\ x+2=-\sqrt5\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\sqrt5-2\\ x=-\sqrt5-2\end{array}\right.\)

b: \(2x^2-4x+1=0\)

=>\(2\left(x^2-2x+\frac12\right)=0\)

=>\(x^2-2x+\frac12=0\)

=>\(x^2-2x+1-\frac12=0\)

=>\(\left(x-1\right)^2=\frac12\)

=>\(\left[\begin{array}{l}x-1=\frac{\sqrt2}{2}\\ x-1=-\frac{\sqrt2}{2}\end{array}\right.\Longrightarrow\left[\begin{array}{l}x=\frac{\sqrt2+2}{2}\\ x=\frac{-\sqrt2+2}{2}\end{array}\right.\)

c: \(\left(2x-1\right)\left(x+2\right)-\left(x-1\right)\left(x-2\right)=2x-9\)

=>\(2x^2+4x-x-2-\left(x^2-3x+2\right)-2x+9=0\)

=>\(2x^2+x+7-x^2+3x-2=0\)

=>\(x^2+4x+5=0\)

=>\(x^2+4x+4+1=0\)

=>\(\left(x+2\right)^2+1=0\) (vô lý)

=>Phương trình vô nghiệm

d: \(\left(x-1\right)\cdot x\cdot\left(x+1\right)\left(x+2\right)=8\)

=>\(x\left(x+1\right)\left(x+2\right)\left(x-1\right)=8\)

=>\(\left(x^2+x\right)\left(x^2+x-2\right)=8\)

=>\(\left(x^2+x\right)^2-2\left(x^2+x\right)-8=0\)

=>\(\left(x^2+x-4\right)\left(x^2+x+2\right)=0\)

mà \(x^2+x+2=x^2+x+\frac14+\frac74=\left(x+\frac12\right)^2+\frac74\ge\frac74>0\forall x\)

nên \(x^2+x-4=0\)

\(\Delta=1^2-4\cdot1\cdot\left(-4\right)=1+16=17>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{-1-\sqrt{17}}{2\cdot1}=\frac{-1-\sqrt{17}}{2}\\ x=\frac{-1+\sqrt{17}}{2\cdot1}=\frac{-1+\sqrt{17}}{2}\end{array}\right.\)

e, Để 5/x <1 thì x<5

\(-2x< 7\Leftrightarrow x>-3,5\) 

\(\left(x-1\right)\left(x-2\right)>0\Leftrightarrow x^2-3x+2>0\Leftrightarrow x^2-3x+\frac{9}{4}>\frac{1}{4}\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2>\frac{1}{4}\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{2}>\frac{1}{2}\\x-\frac{3}{2}< -\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>2\\x< 1\end{cases}}\)

a) \(2x\left(x-\frac{1}{7}\right)=0\)

\(x\left(x-\frac{1}{7}\right)=0\)

\(\Rightarrow2x-2.\frac{1}{7}=0\)

\(2x-\frac{2}{7}=0\)

=> \(2x=\frac{2}{7}\)

=> x=\(\frac{1}{7}\)

b) (x-9)(\(x+\frac{3}{5}\))=0

\(\Rightarrow\orbr{\begin{cases}x-9=0\\x+\frac{3}{5}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-3}{5}\end{cases}}\)

Vậy x=0 hoặc x=-3/5

c) \(\left(\frac{-4}{7}-2x\right)\left(x-\frac{5}{4}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{-4}{7}-2x=0\\x-\frac{5}{4}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-2}{7}\\x=\frac{5}{4}\end{cases}}\)

Vậy x=-2/7 hoặc x=5/4

a, => x.(x-1/7) = 0:2 = 0

=> x=0 hoặc x-1/7=0

=> x=0 hoặc x=1/7

Vậy x thuộc {0;1/7}

b, => x-9=0 hoặc x+3/5=0

=> x=9 hoặc x=-3/5

Vậy x thuộc {-3/5;9}

c, => -4/7-2x=0 hoặc x-5/4=0

=> x=-2/7 hoặc x=5/4

Vậy x thuộc {-2/7;5/4}

Tk mk nha

a: ta có: \(\frac{x}{y}=\frac34\)

=>\(\frac{x}{3}=\frac{y}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{3}=\frac{y}{4}=\frac{2x+5y}{2\cdot3+5\cdot4}=\frac{52}{6+20}=\frac{52}{26}=2\)

=>\(\begin{cases}x=2\cdot3=6\\ y=2\cdot4=8\end{cases}\)

b: \(\frac{2x}{3y}=-\frac13\)

=>-6x=3y

=>\(-2x=y\)

2x+5y=9

=>\(2x+5\cdot\left(-2x\right)=9\)

=>2x-10x=9

=>-8x=9

=>\(x=-\frac98\)

\(y=-2x=-2\cdot\frac{-9}{8}=\frac94\)

c: 21x=9y

=>7x=3y

=>\(\frac{x}{3}=\frac{y}{7}\)

mà x-y=24

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{3}=\frac{y}{7}=\frac{x-y}{3-7}=\frac{24}{-4}=-6\)

=>\(\begin{cases}x=-6\cdot3=-18\\ y=-6\cdot7=-42\end{cases}\)

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