a. 

  150 + x : 3 = 620 : 4 

   150 + x : 3 = 155

            x : 3 = 155 - 150

            x : 3 = 5 

            x     = 5 x 3 

           x     = 15 

b.

   x * 0,1 = 1/2 -2/5

   x * 1/10 = 5/10 - 4/10 

   x * 1/10 = 1/10

  x            = 1/10 : 1/10 

  x            = 1 

c.

4/9 + 5/9 : x = 1 

        5/9 : x = 1 - 4/9

        5/9 :x  = 5/9

              x  = 5/9 : 5/9

              x  = 1 

1)  \(150+x.\frac{1}{3}=155\)

\(\frac{1}{3}x=5\)

\(x=15\)

2)  \(x.\frac{1}{10}=\frac{1}{2}-\frac{2}{5}\)

\(\frac{1}{10}x=\frac{5}{10}-\frac{4}{10}\)

\(\frac{1}{10}x=\frac{1}{10}\)

\(x=1\)

3)  \(\frac{4}{9}+\frac{9}{5}x=1\)

\(\frac{9}{5}x=\frac{9}{9}-\frac{4}{9}\)

\(\frac{9}{5}x=\frac{5}{9}\)

\(x=1\)

Tính phải k nhỉ?

`1)`

`2x + 3x + 5x`

`= (2 + 3 + 5)x`

`= 10x`

`2)`

`2.x - x + 3.x`

`= (2 - 1 + 3)x`

`= 4x`

`3)`

`9.x - 3 - 3.x`

`= (9 - 3)x - 3`

`= 6x - 3`

`4)`

Thiếu dấu, bạn bổ sung thêm

`5)`

`x - 0,2x - 0,1x`

`= (1 - 0,2 - 0,1)x`

`=0,7x`

`6)`

\(\dfrac{7}{2}x-\dfrac{1}{2}x=\left(\dfrac{7}{2}-\dfrac{1}{2}\right)x=3x\)

a. Vì \(0< 0,1< 1\) nên bất phương trình đã cho 

\(\Leftrightarrow0< x^2+x-2< x+3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x-2>0\\x^2-5< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< -2\\x>1\end{matrix}\right.\\-\sqrt{5}< x< \sqrt{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{5}< x< -2\\1< x< \sqrt{5}\end{matrix}\right.\)

Vậy tập nghiệm của bất phương trình là \(S=\left\{-\sqrt{5};-2\right\}\) và \(\left\{1;\sqrt{5}\right\}\)

b. Điều kiện \(\left\{{}\begin{matrix}2-x>0\\x^2-6x+5>0\end{matrix}\right.\)

Ta có:

 \(log_{\dfrac{1}{3}}\left(x^2-6x+5\right)+2log^3\left(2-x\right)\ge0\)

\(\Leftrightarrow log_{\dfrac{1}{3}}\left(x^2-6x+5\right)\ge log_{\dfrac{1}{3}}\left(2-x\right)^2\)

\(\Leftrightarrow x^2-6x+5\le\left(2-x\right)^2\)

\(\Leftrightarrow2x-1\ge0\)

Bất phương trình tương đương với:

\(\left\{{}\begin{matrix}x^2-6x+5>0\\2-x>0\\2x-1\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1\\x>5\end{matrix}\right.\\x< 2\\x\ge\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{1}{2}\le x< 1\)

Vậy tập nghiệm của bất phương trình là: \(\left(\dfrac{1}{2};1\right)\)

Lời giải:

a. $x(\frac{4}{5}x-1)(0,1x-10)=0$

$\Rightarrow x=0$ hoặc $\frac{4}{5}x-1=0$ hoặc $0,1x-10=0$

Nếu $\frac{4}{5}x-1=0$

$\Rightarrow x=1: \frac{4}{5}=\frac{5}{4}$

Nếu $0,1x-10=0$

$\Rightarrow x=10:0,1=100$

Vậy $x=0; \frac{5}{4}; 100$

b.

$(\frac{1}{4}x-1)-(\frac{5}{6}x+2)-(1-\frac{5}{8}x)=0$

$(\frac{1}{4}x-\frac{5}{6}x+\frac{5}{8}x)-(1+2+1)=0$

$\frac{1}{24}x-4=0$

$x=4: \frac{1}{24}=96$

\(\Leftrightarrow x\cdot\dfrac{21}{2}=\dfrac{3}{7}\)

hay x=2/49

a: Ta có: \(8x+11-3=5x+x-3\)

\(\Leftrightarrow8x+8=6x-3\)

\(\Leftrightarrow2x=-11\)

hay \(x=-\dfrac{11}{2}\)

b: Ta có: \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)

\(\Leftrightarrow2x\left(x^3+6x^2+12x+8\right)-8x^2=2\left(x^3-8\right)\)

\(\Leftrightarrow2x^4+12x^3+24x^2+16x-8x^2-2x^3+16=0\)

\(\Leftrightarrow2x^4+10x^3+16x^2+16x+16=0\)

\(\Leftrightarrow2x^4+4x^3+6x^3+12x^2+4x^2+8x+8x+16=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x^3+6x^2+4x+8\right)=0\)

\(\Leftrightarrow x+2=0\)

hay x=-2

c: Ta có: \(\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\)

\(\Leftrightarrow2x^2-3x+2x-3-2x^2-10x+x+5=0\)

\(\Leftrightarrow-10x+2=0\)

\(\Leftrightarrow-10x=-2\)

hay \(x=\dfrac{1}{5}\)

d: Ta có: \(\dfrac{1}{10}-2\cdot\left(\dfrac{1}{2}t-\dfrac{1}{10}\right)=2\left(t-\dfrac{5}{2}\right)-\dfrac{7}{10}\)

\(\Leftrightarrow\dfrac{1}{10}-t+\dfrac{1}{5}=2t-5-\dfrac{7}{10}\)

\(\Leftrightarrow-t-2t=-\dfrac{57}{10}-\dfrac{3}{10}=-6\)

hay t=2

a: \(\left(\sqrt{3}\right)^x=243\)

=>\(3^{\dfrac{1}{2}\cdot x}=3^5\)

=>\(\dfrac{1}{2}\cdot x=5\)

=>x=10

b: \(0,1^x=1000\)

=>\(\left(\dfrac{1}{10}\right)^x=1000\)

=>\(10^{-x}=10^3\)

=>-x=3

=>x=-3

c: \(\left(0,2\right)^{x+3}< \dfrac{1}{5}\)

=>\(\left(0,2\right)^{x+3}< 0,2\)

=>x+3>1

=>x>-2

d: \(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{5}{3}\right)^2\)

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

=>2x+1<-2

=>2x<-3

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

e: \(5^{x-1}+5^{x+2}=3\)

=>\(5^x\cdot\dfrac{1}{5}+5^x\cdot25=3\)

=>\(5^x=\dfrac{3}{25,2}=\dfrac{1}{8,4}=\dfrac{10}{84}=\dfrac{5}{42}\)

=>\(x=log_5\left(\dfrac{5}{42}\right)=1-log_542\)

\(a,\dfrac{1}{5}+\dfrac{1}{2}+\dfrac{3}{10}=\dfrac{2+5+3}{10}=\dfrac{10}{10}=1\\ 3+\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{3\times6+2+3+1}{6}=4\\ Vậy:1< x< 4\\ Vậy:x=2.hoặc.x=3\)

\(b,1000\times0,1=100\\ 10,3:0,1=103\\ Vậy:100< x< 103\\ Vậy:x=101.hoặc.x=102\)

A 3x-4x=-9-3

    -x=-12

     x=12

B 3.2x -5x +1=5+0.2x

  3.2x-5x-0.2x=5-1

  -2x=4

 x=-2

C 1.5-x-2=-3x-0.3

  -x+3x=-0.3-1.5+2

  2x =0.2

  x=0.1

E 2/3-1/2x-1=-x+1

  -1/2x+x=1+1-2/3

  1/2x=4/3

  x=8/3

F 3t-4+13+2t+4-3t

  =3t+2t-3t-4+13+4

  =2t+13