1.

\(\left(\dfrac{-1}{8}+\dfrac{-5}{6}\right)\cdot\dfrac{6}{23}\\ =-\dfrac{23}{24}\cdot\dfrac{6}{23}\\ =-\dfrac{6}{24}=-\dfrac{1}{4}\)

2. Xem lại đề nha!

4.

\(x+0,75=-1\dfrac{1}{4}\\ x+\dfrac{3}{4}=-\dfrac{3}{4}\\ x=-\dfrac{3}{4}-\dfrac{3}{4}\\ x=-\dfrac{3}{4}+\left(-\dfrac{3}{4}\right)=-\dfrac{6}{4}=-\dfrac{3}{2}\)

5.

\(\dfrac{x}{28}=-\dfrac{4}{7}\\ \Leftrightarrow7x=-4.28\\ \Rightarrow7x=-112\\ \Rightarrow x=-112:7=-16\)

6.

\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Leftrightarrow\left(3x-y\right).4=3\left(x+y\right)\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow12x-3x=4y+3y\\ \Rightarrow9x=7y\\ \Leftrightarrow\dfrac{x}{7}=\dfrac{y}{9}\Leftrightarrow\dfrac{x}{y}=\dfrac{7}{9}\)

Vậy giá trị của tỉ số \(\dfrac{x}{y}=\dfrac{7}{9}\).

trả lời nhanh giúp mình nha

CẢM ƠN CÁC BẠN

b) x/-4=-48/3x

=>x.3x=-4.(-48)

=>3x²= 192

=>x²=64

=>x=8

Vậy............

\(a,1\dfrac{2}{3}x-\dfrac{1}{4}=\dfrac{5}{6}.\)

\(1\dfrac{2}{3}x=\dfrac{5}{6}+\dfrac{1}{4}.\)

\(1\dfrac{2}{3}x=\dfrac{13}{12}.\)

\(x=\dfrac{13}{12}:1\dfrac{2}{3}.\)

\(x=\dfrac{13}{20}.\)

Vậy \(x=\dfrac{13}{20}.\)

\(c,3^{2x+1}=81.\)

\(3^{2x+1}=3^4.\)

\(\Rightarrow2x+1=4.\)

\(\Rightarrow2x=5.\)

\(\Rightarrow x=\dfrac{5}{2}.\)

Vậy \(x=\dfrac{5}{2}.\)

Các câu dễ tự làm :v

\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)

\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)

\(\Rightarrow x+1=0\Rightarrow x=-1\)

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Rightarrow x+2004=0\Rightarrow x=-2004\)

1: \(\Leftrightarrow\left(x+1\right)^2=4\)

=>x+1=2 hoặc x+1=-2

=>x=1 hoặc x=-3

2: \(\Leftrightarrow7x-21=5x+25\)

=>2x=46

=>x=23

3: \(\Leftrightarrow x^2+4x+3=x^2+0.5x+4x+2\)

=>4,5x+2=4x+3

=>x=1

mn ơi giúp nhé

a)A=\(x^5-\dfrac{1}{2}x+7x^3-2x+\dfrac{1}{5}x^3+3x^4-x^5+\dfrac{2}{5}x^4+15\)

=\(=\dfrac{-5}{2}x+\dfrac{36}{5}x^3+\dfrac{17}{5}x^4+15\)

b)B=\(3x^2-10+\dfrac{2}{5}x^3+7x-x^2+8+7x^2\)

\(=9x^2+\dfrac{2}{5}x^3+7x+2\)

c)C=\(\dfrac{1}{7}x-2x^4+5x+6\)

c)C=\(\dfrac{36}{7}x-2x^4+6\)

a) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)

\(20x^2-16x-34=10x^2+3x-34\)

\(10x^2-19x=0\)

\(x\left(10x-19\right)=0\)

\(\Leftrightarrow x=0\)

hoặc \(10x-19=0\)

\(\Leftrightarrow x=\dfrac{19}{10}\)

Vạy ..............

b) \(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)

\(\Leftrightarrow1-\dfrac{x-1}{x+5}=1-\dfrac{6}{7}\)

\(\Leftrightarrow\dfrac{x+5}{x+5}-\dfrac{x-1}{x+5}=\dfrac{7}{7}-\dfrac{6}{7}\)

\(\Leftrightarrow\dfrac{\left(x+5\right)-\left(x-1\right)}{x+5}=\dfrac{1}{7}\)

\(\Leftrightarrow\dfrac{x+5-x+1}{x+5}=\dfrac{1}{7}\)

\(\Leftrightarrow\dfrac{\left(x-x\right)+\left(5+1\right)}{x+5}=\dfrac{1}{7}\)

\(\Leftrightarrow\dfrac{6}{x+5}=\dfrac{1}{7}\)

\(\Leftrightarrow x+5=42\)

\(\Leftrightarrow x=37\)

c) x = 0 và x = -1/2