mình có đáp án rồi ạ.

`@` `\text {Ans}`

`\downarrow`

`a)`

`2^2 * 16 \ge 2^x \ge 4^2`

`=> 2^2 * 2^4 \ge 2^x \ge 2^4`

`=> 2^6 \ge 2^x \ge 2^4`

`=> x \in {4; 5; 6}`

`b)`

`9*27 \le 3^x \le 243`

`=> 3^2 * 3^3 \le 3^x \le 3^5`

`=> 3^5 \le 3^x \le 3^5`

`=> x = 5`

`c)`

`2 * (x - 1/2)^2 - 1/8 = 0`

`=> 2* (x - 1/2)^2 = 1/8`

`=> (x - 1/2)^2 = 1/8 \div 2`

`=> (x-1/2)^2 = 1/16`

`=> (x - 1/2)^2 = (+- 1/4)^2`

`=>`\(\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{1}{4}\\x-\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{4}+\dfrac{1}{2}\\x=\dfrac{1}{2}-\dfrac{1}{4}\end{matrix}\right.\)

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

Vậy, `x \in {1/4; 3/4}.`

a) \(\dfrac{x+1}{4}-\dfrac{5+2x}{8}=\dfrac{3-4x}{2}\)

⇔\(\dfrac{2\left(x+1\right)}{8}-\dfrac{5+2x}{8}=\dfrac{4\left(3-4x\right)}{8}\) 

⇔ 2x + 2 - 5 - 2x = 12 -16x

⇔ 16x = 15 

⇔ x = 15/16

b) \(\dfrac{4-3x}{5}-\dfrac{4-x}{10}=\dfrac{x+2}{2}\)

⇔\(\dfrac{2\left(4-3x\right)}{10}-\dfrac{4-x}{10}=\dfrac{5\left(x+2\right)}{10}\)

⇔ 8 - 6x - 4 + x = 5x + 10

⇔ 10x = -6

⇔ x = -6/10

Câu 1:

x + 1/4 - 5 + 2x/8 = 3 - 4x/2

<=> 2x + 2/8 - 5 + 2x/8 = 12 - 16x/8

<=> 2x + 2 - 5 - 2x = 12 - 16x

<=> -3 = 12 - 16x <=> 15 = 16x <=> x = 15/16

Câu 2:

4 - 3x/5 - 4 - x/10 = x + 2/2

<=> 8 - 6x/10 - 4 - x/10 = 5x + 10/10

<=> 8 - 6x - 4 + x = 5x + 10

<=> 4 - 5x = 5x + 10

<=> 4 = 10x + 10 <=> 10x = -6 <=> x = -3/5

Bài này dễ nên nãy không có định làm, nhưng mà nghĩ lại thôi làm vậy:Đ

a/ \(\dfrac{x}{-2}=\dfrac{-4}{y}=\dfrac{2}{4}\)

Ta có: \(\dfrac{-4}{y}=\dfrac{2}{4}\Rightarrow y=\dfrac{-4.4}{2}=-8\)

\(\dfrac{x}{-2}=\dfrac{-4}{y}=\dfrac{-4}{-8}=\dfrac{1}{2}\Rightarrow x=\dfrac{-2.1}{2}=-1\)

b/\(\dfrac{2}{x}=\dfrac{y}{-3}\Rightarrow xy=-3.2=-6\)

\(\Rightarrow x;y\inƯ\left(-6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Ta có bảng giá trị của x, y như sau:

c/ \(\dfrac{x+1}{2}=\dfrac{8}{x+1}\Rightarrow\left(x+1\right)^2=2.8=16\)

\(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)^2=4^2\\\left(x+1\right)^2=\left(-4\right)^2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

c,(x+1).(x+1)=8.2

(x+1)²=16

⇒x+1=4

⇒x=3

a: \(P=\dfrac{8+5x-2x-8}{x\left(x+4\right)}=\dfrac{3x}{x\left(x+4\right)}=\dfrac{3}{x+4}\)

b: Khi x=1/2 thì P=3/(1/2+4)=3:9/2=3*2/9=6/9=2/3

\(a,\left(\dfrac{31}{35}-\dfrac{4}{7}\right)\times\dfrac{8}{7}:2\\ =\left(\dfrac{31}{35}-\dfrac{4\times5}{35}\right)\times\dfrac{8}{7}:2\\ =\dfrac{11}{35}\times\dfrac{8}{7}:2\\ =\dfrac{88}{245}:2\\ =\dfrac{44}{245}\\ b,\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\\ =\left(\dfrac{2-1}{2}\right)\times\left(\dfrac{3-1}{3}\right)\times\left(\dfrac{4-1}{4}\right)\times\left(\dfrac{5-1}{5}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)

a, ( \(\dfrac{31}{35}\) - \(\dfrac{4}{7}\)) \(\times\) \(\dfrac{8}{7}\): 2

= \(\left(\dfrac{31}{35}-\dfrac{20}{35}\right)\) \(\times\) \(\dfrac{8}{7}\) : 2

= \(\dfrac{11}{35}\) \(\times\) \(\dfrac{8}{7}\) \(\times\) \(\dfrac{1}{2}\)

= \(\dfrac{44}{35}\) \(\times\) \(\dfrac{4}{7}\)

= \(\dfrac{44}{245}\)

b, ( 1 - \(\dfrac{1}{2}\)) \(\times\) ( 1 - \(\dfrac{1}{3}\)) \(\times\) ( 1 - \(\dfrac{1}{4}\)) \(\times\) ( 1 - \(\dfrac{1}{5}\))

= \(\dfrac{1}{2}\) \(\times\) \(\dfrac{2}{3}\) \(\times\) \(\dfrac{3}{4}\) \(\times\) \(\dfrac{4}{5}\)

= \(\dfrac{1}{5}\) \(\times\) \(\dfrac{2\times3\times4}{2\times3\times4}\)

= \(\dfrac{1}{5}\)

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10

ý bạn là tìm x hay sao?

\(a,\Leftrightarrow\dfrac{\left(x-2\right)\left(x+1\right)}{x+1}=\dfrac{x^2-3x-2}{x-1}\left(x\ne\pm1\right)\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=x^2-3x-2\\ \Leftrightarrow x^2-3x+2=x^2-3x-2\\ \Leftrightarrow2=-2\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{x^2-2x+4}=x+2\\ \Leftrightarrow x+2=x+2\\ \Leftrightarrow x\in R\)

 alo cho tui hỏi bạn có phải Dương Ngọc Lan Hương Trường THCS Minh Thuận 3 k dọ