A) x = 5

B) x = 18

C) thì ko bít

a. \(\dfrac{x}{5}=\dfrac{25}{75}\)

\(x:5=25:75=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}\text{×}5=\dfrac{5}{3}\)

b. \(\dfrac{16}{x}=\dfrac{8}{9}\)

\(16:x=8:9\)

\(x=16:8:9=\dfrac{2}{9}\)

c. \(\dfrac{20}{35}=\dfrac{x}{14}\)

\(x:14=20:35=\dfrac{4}{7}\)

\(x=\dfrac{4}{7}\text{×}14=8\)

Lời giải:
a.

$x=\frac{25}{75}\times 5=\frac{25\times 5}{75}=\frac{25\times 5}{25\times 3}=\frac{5}{3}$

b.

$x=16: \frac{8}{9}=16\times \frac{9}{8}=\frac{8\times 2\times 9}{8}=18$

c.

$x=\frac{20}{35}\times 14=\frac{20\times 14}{35}=\frac{4\times 5\times 7\times 2}{5\times 7}=4\times 2=8$

\(\dfrac{1}{x}+\dfrac{3}{2x}-\dfrac{5}{24}=0\)

\(\Leftrightarrow24+36-5x=0\)

\(\Leftrightarrow5x=60\)

\(\Leftrightarrow x=12\)

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

\(\Leftrightarrow2x-10+3x+3>18\)

\(\Leftrightarrow5x>25\)

\(\Leftrightarrow x>5\)

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

\(\Leftrightarrow2x+4+x-1=2x\)

\(\Leftrightarrow x=-3\)

Nửa chu vi hcn là : 5/2 : 2 = 5/4

Chiều dài hcn là : ( 5/4 + 1/4 ) : 2 = 3/4

Chiều rộng hcn là : 5/4 - 3/4 = 1/2

Nửa chu vi là:

5/2:2=5/4(m)

Chiều dài là:

(5/4+1/4):2=3/4(m)

Chiều rộng là:

5/4-3/4=1/2(m)

ĐS:...

Nửa chu vi hcn là:
\(\dfrac{5}{2}:2=\dfrac{5}{4}\left(m\right)\)

Chiều dài hcn là:
\(\left(\dfrac{5}{4}+\dfrac{1}{4}\right):2=\dfrac{3}{4}\left(m\right)\)

Chiều rộng hcn là:

\(\dfrac{5}{4}-\dfrac{3}{4}=\dfrac{1}{2}\left(m\right)\)

Bài 4:

a: Ta có: \(-\left|x+1.1\right|\le0\forall x\)

\(\Leftrightarrow-\left|x+1.1\right|+1.5\le1.5\forall x\)

Dấu '=' xảy ra khi x=-1,1

b: Ta có: \(-4\left|x-2\right|\le0\forall x\)

\(\Leftrightarrow-4\left|x-2\right|+10\le10\forall x\)

Dấu '=' xảy ra khi x=2

24/12

24/12

\(a,A=0,2\left(5x-1\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x+4\right)+\dfrac{2}{3}\left(3-x\right)\)

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

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

\(=-0,2\)

\(b,B=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x^3-8y^3+10\right)\)

\(=x^3-8y^3-x^3+8y^3-10\)

\(=-10\)

\(c,C=4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)-4x\)

\(=4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-4x\)

\(=4x^2+8x+4+4x^2-4x+1-8x^2+8-4x\)

\(=13\)

a) \(A=0,2\left(5x-1\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x+4\right)+\dfrac{2}{3}\left(3-x\right)\)

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

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

\(A=-\dfrac{1}{5}\)

Vậy: ...

b) \(B=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x^3-8y^3+10\right)\)

\(B=\left[x^3-\left(2y\right)^3\right]-\left[x^3-\left(2y\right)^3\right]-10\)

\(B=-10\)

Vậy: ...

c) \(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x+1\right)\left(x-1\right)-4x\)

\(=4\left(x^2+2x+4\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-4x\)

\(=4x^2+8x+4+4x^2-4x+1-8x^2+8-4x\)

\(=\left(4x^2+4x^2-8x^2\right)+\left(8x-4x-4x\right)+\left(4+1+8\right)\)

\(=13\)

Vậy:...

=x−0,2−13x−2+2−23x=�−0,2−13�−2+2−23�

1. A

2. C

3. B

4. A

5. C

6. D