b: x=-3/4

x= +- 2,1

x= - 17/9

x= +- 1/2/5

x= 0,35

\(\left|x\right|=2,1\)

\(\Rightarrow x=+_-2,1\)

\(\left|x\right|=\frac{17}{9}vax< 0\)

\(\Rightarrow x=-\frac{17}{9}\)

\(\left|x\right|=\frac{1}{\frac{2}{5}}=\frac{5}{2}\)

\(\Rightarrow x=+_-2,5\)

\(\left|x\right|=0,35vax>0\)

\(\Rightarrow x=0,35\)

Câu 1:

a)|x|=2,1

Suy ra:\(x=\frac{21}{10};-\frac{21}{10}\)

b)|x|=1

    |x|=\(\frac{2}{5}\)

TH1:x có dạng \(\frac{a}{a};-\frac{a}{a}\)(a thuộc mọi điều kiện)

TH2:\(x=\frac{2}{5};-\frac{2}{5}\)

c)|x|=\(\frac{17}{9}\);x<0

TH1:\(x=\frac{17}{9};-\frac{17}{9}\)

TH2:Vì ko có giá trị tuyệt đối nào nhỏ hơn ko

Suy ra x thuộc tập rỗng

d)|x|=0,35 và x>0

TH1:\(x=\frac{7}{20};-\frac{7}{20}\)

TH2:Vì x>0 suy ra x thuộc mọi điều kiền (trừ số 0)

Câu 2:

a)|x-1,7|=2,3

Suy ra:

TH1:x-1,7=2,3

        x=4

TH2:x-1,7=-2,3

        x=-0,6

                Vậy x=4;-0,6

b)\(\left|x+\frac{3}{4}\right|-\frac{1}{3}=0\)

    \(\left|x+\frac{3}{4}\right|=\frac{1}{3}\)

TH1:\(x+\frac{3}{4}=\frac{1}{3}\)

        \(x=-\frac{5}{12}\)

TH2:\(x+\frac{3}{4}=-\frac{1}{3}\)

        \(x=-\frac{13}{12}\)

   Vậy \(x=-\frac{5}{12}\);\(x=-\frac{13}{12}\)

a) x =\(\dfrac{-3}{4}\)

\(\left|x\right|=\frac{3}{4}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{-3}{4}\end{cases}}\)

mà \(x< 0\)nên \(x=\frac{-3}{4}\)

vậy \(x=\frac{-3}{4}\)

\(\left|x\right|=0,35\)

\(\left|x\right|=\frac{7}{20}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{20}\\x=\frac{-7}{20}\end{cases}}\)

mà \(x>0\)nên \(x=\frac{7}{20}\)

vậy \(x=\frac{7}{20}\)

|x|=3/4 => x=3/4 hoặc x= -3/4

|x|=0,35 => x=0,35 hoặc x= -0,35

1 ) \(\left(x-4\right)^2-25=0\)

\(\Leftrightarrow\left(x-4-5\right)\left(x-4+5\right)=0\)

\(\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)

2 ) \(\left(x-3\right)^2-\left(x-1\right)^2=0\)

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

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

\(\Leftrightarrow x=2.\)

3 ) \(\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\)

\(\Leftrightarrow\left(x^2-4\right)\left(2x+3-x+1\right)=0\)

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

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

4 ) \(\left(x^2-1\right)-\left(x+1\right)\left(2-3x\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-1-2+3x\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(4x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{3}{4}\end{matrix}\right.\)

5 ) \(x^3+x^2+x+1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=-1.\end{matrix}\right.\)

6 ) \(x^3+x^2-x-1=0\)

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

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

7 ) \(2x^3+3x^2+6x+5=0\)

\(\Leftrightarrow2x^3+2x^2+x^2+x+5x+5=0\)

\(\Leftrightarrow2x^2\left(x+1\right)+x\left(x+1\right)+5\left(x+1\right)=0\)

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

\(\Leftrightarrow x=-1.\)

8 ) \(x^4-4x^3-19x^2+106x-120=0\)

\(\Leftrightarrow x^4-4x^3-19x^2+76x+30x-120=0\)

\(\Leftrightarrow x^3\left(x-4\right)-19x\left(x-4\right)+30\left(x-4\right)=0\)

\(\Leftrightarrow\left(x^3-19x+30\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left(x^3-8-19x+38\right)\left(x-4\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+4x+23\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

9 ) \(\left(x^2-3x+2\right)\left(x^2+15x+56\right)+8=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+7\right)\left(x+8\right)+8=0\)

\(\Leftrightarrow\left(x^2+7x-x-7\right)\left(x^2+8x-2x-16\right)+8=0\)

\(\Leftrightarrow\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8=0\)

Đặt \(x^2+6x-7=t\)

\(\Leftrightarrow t\left(t-9\right)+8=0\)

\(\Leftrightarrow t^2-9t+8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=8\\t=1\end{matrix}\right.\)

Khi t = 8 \(\Leftrightarrow x^2+6x-7=8\Leftrightarrow x^2+6x-15\Leftrightarrow\left[{}\begin{matrix}x=-3+2\sqrt{6}\\x=-3-2\sqrt{6}\end{matrix}\right.\)

Khi t = 1 \(\Leftrightarrow x^2+6x-7=1\Leftrightarrow x^2+6x-8=0\Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{17}\\x=-3-\sqrt{17}\end{matrix}\right.\)

Vậy ........

Bài 1:

\(a)\left(\dfrac{-28}{29}\right).\left(\dfrac{-38}{16}\right)=\dfrac{\left(-28\right).\left(-38\right)}{29.16}=\dfrac{1064}{464}=\dfrac{133}{58}\)

\(b)\left(\dfrac{-21}{16}\right).\left(\dfrac{-24}{7}\right)=\dfrac{\left(-21\right).\left(-24\right)}{16.7}=\dfrac{504}{112}=\dfrac{9}{2}\)

\(c)\left|\dfrac{-12}{17}\right|.\left(\dfrac{-34}{9}\right)=\dfrac{12}{17}.\left(\dfrac{-34}{9}\right)=\dfrac{12.\left(-34\right)}{17.9}=\dfrac{-408}{153}=\dfrac{-8}{3}\)

Bài 3:

\(a)\left|x\right|=21\)

\(\Rightarrow\left[{}\begin{matrix}x=-21\\x=21\end{matrix}\right.\)

\(b)\left|x\right|=\dfrac{17}{9};x< 0\)

\(\Rightarrow x=\dfrac{-17}{9}\)

\(c)\left|x\right|=1\)

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

\(\left|x\right|=\dfrac{2}{5}\)

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

\(d)\left|x\right|=0,35;x>0\)

\(\Rightarrow x=0,35\)

Bài 4:

\(a)\left|x\right|-1,7=2,3\)

\(\Rightarrow\left[{}\begin{matrix}x-1,7=2,3\\x-1,7=-2,3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{-3}{5}\end{matrix}\right.\)

\(b)\left|x\right|+\dfrac{3}{4}-\dfrac{1}{3}=0\)

\(\Rightarrow\left|x\right|+\dfrac{3}{4}=0+\dfrac{1}{3}\)

\(\Rightarrow\left|x\right|+\dfrac{3}{4}=\dfrac{1}{3}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{3}\\x+\dfrac{3}{4}=\dfrac{-1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-5}{12}\\x=\dfrac{-13}{12}\end{matrix}\right.\)

Chúc bạn học tốt!

1,-12(x-5)+7(3-x)=5

=>-12x+60+21-7x=5

=>-12x-7x+60+21=5

=>-19x+81=5

=>-19x=5-81

=>-19x=-76

=>x=(-76):(-19)

=>x=4

2,(x-2) (x+4) =0

=>+,x-2=0 => x=2

+,x+4=0 => x=-4

Vậy x=2 hoặc x=-4

3,(x-2) (x+15) =0

=>+,x-2=0 =>x=2

+,x+15=0 =>x=-15

Vậy x=2 hoặc x=-15

4,(7-x) (x+19) =0

=>+,7-x=0 =>x=7

+,x+19=0 =>x=-19

Vậy x=7 hoặc x=-19

5,(x-3) (x-5)<0

=>x-3 và x-5 là hai số khác dấu

TH1

+,x-3<0 =>x<3(1)

+,x-5>0 =>x>5 (2)

Từ (1) và(2) => 5<x<3(Vô lí nên trường hợp này bị loại)

TH2

+,x-3>0 =>x>3 (3)

+,x-5<0 =>x<5 (4)

Từ (3) và (4) =>3<x<5 => x=4

Vậy x=4

Chú bn hc tốt hơn nha!!