a: x<5 thì 5-x>0

A=5x+5-x+5=4x+10

b: Khi x>=0 thì \(B=5x+10+3x=8x+10\)

Khi x<0 thì B=5x+10-3x=2x+10

d: Khi x>=3 thì \(D=x-3-3x+15=-2x+12\)

Khi x<3 thì D=3-x-3x+15=-4x+18

1) 4x-(2x-5)=21

4x-2x+5=21

2x+5=21

2x=21 -5

2x=16

x=16/2

x=8

2)14x+5=8x+10

14x-8x=10-5

6x=5

x=5/6

3) 15+5x=3x+30

5x-3x=30-15

2x=15

x=15/2

Vậy x=15/2

1) 14x-8x=10+5

x(14-8)=15

x6=15

x=15/6

2)5x-3x=30-15

2x=15

x=15/2

3)làm tương tự

1) x=2,5

2) x=7,5

3) x=4

4) x=7/3

5) x=8,25

a) Ta có: \(3x\left(7x-2\right)-14x+4=0\)

\(\Leftrightarrow3x\left(7x-2\right)-2\left(7x-2\right)=0\)

\(\Leftrightarrow\left(7x-2\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}7x-2=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=2\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{7}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{2}{7};\dfrac{2}{3}\right\}\)

b) ĐKXĐ: \(x\notin\left\{0;3\right\}\)

Ta có: \(\dfrac{2x+1}{x-3}+\dfrac{5-3x}{x}=\dfrac{2x^2-15}{x^2-3x}\)

\(\Leftrightarrow\dfrac{x\left(2x+1\right)}{x\left(x-3\right)}+\dfrac{\left(5-3x\right)\left(x-3\right)}{x\left(x-3\right)}=\dfrac{2x^2-15}{x\left(x-3\right)}\)

Suy ra: \(2x^2+x+5x-15-3x^2+9x-2x^2+15=0\)

\(\Leftrightarrow-3x^2+15x=0\)

\(\Leftrightarrow-3x\left(x-5\right)=0\)

mà -3<0

nên x(x-5)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=5\left(nhận\right)\end{matrix}\right.\)

Vậy: S={5}

b) PT \(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow\left(15x-35\right)\left(5x+3\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

 Vậy \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)

c) PT \(\Leftrightarrow\left(2-3x\right)\left(x-11\right)+\left(2-3x\right)\left(2-5x\right)=0\)

\(\Leftrightarrow\left(2-3x\right)\left(-9-4x\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{9}{4}\end{matrix}\right.\)

  Vậy \(S=\left\{\dfrac{2}{3};-\dfrac{9}{4}\right\}\)

a)(x-1)(5x+3)=(3x-8)(x-1)

\(\Leftrightarrow\)(x-1)(5x+3)-(3x-8)(x-1)=0

\(\Leftrightarrow\left(x-1\right)\left(5x-3-3x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\)

\(\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{1;\dfrac{5}{2}\right\}\)

a) Ta có: \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

\(\Leftrightarrow5x^2+3x-5x-3=3x^2-3x-8x+8\)

\(\Leftrightarrow5x^2-2x-3=3x^2-11x+8\)

\(\Leftrightarrow5x^2-2x-3-3x^2+11x-8=0\)

\(\Leftrightarrow2x^2+9x-11=0\)

\(\Leftrightarrow2x^2+11x-2x-11=0\)

\(\Leftrightarrow x\left(2x+11\right)-\left(2x+11\right)=0\)

\(\Leftrightarrow\left(2x+11\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+11=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-11\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{11}{2}\\x=1\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{11}{2};1\right\}\)

b) Ta có: \(3x\left(25x+15\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow3x\cdot5\cdot\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow\left(5x+3\right)\left(15x-35\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+3=0\\15x-35=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\15x=35\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=\dfrac{7}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)

c) Ta có: \(\left(2-3x\right)\left(x-11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\Leftrightarrow2x-22-3x^2+33x=6x-15x^2-4+10x\)

\(\Leftrightarrow-3x^2+35x-22=-15x^2+16x-4\)

\(\Leftrightarrow-3x^2+35x-22+15x^2-16x+4=0\)

\(\Leftrightarrow12x^2+19x-18=0\)

\(\Leftrightarrow12x^2+27x-8x-18=0\)

\(\Leftrightarrow3x\left(4x+9\right)-2\left(4x+9\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}4x+9=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-9\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{4}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{9}{4};\dfrac{2}{3}\right\}\)

a)\(\left(x-15\right).3=125\)

\(x-5=125:3\)

\(x-5=41,66666666666667\)

\(x=41,66666666666667+5\)

\(x=46,66666666666667\)

b) \(2.x+3=17\)

\(2.x=17-3\)

\(2.x=14\)

\(x=14:2\)

\(x=7\)

c) \(15-3.x=3\)

\(3.x=15-3\)

\(3.x=12\)

\(x=12:3\)

\(x=4\)

K mình nha    Monkey Kiều Anh

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