\(\frac{x+1}{7}+\frac{x+2}{6}=\frac{x+3}{5}+\frac{x+4}{4}\)

\(\Leftrightarrow\left(\frac{x+1}{7}+1\right)+\left(\frac{x+2}{6}+1\right)=\left(\frac{x+3}{5}+1\right)+\left(\frac{x+4}{4}+1\right)\)

\(\Leftrightarrow\frac{x+8}{7}+\frac{x+8}{6}-\frac{x+8}{5}-\frac{x+8}{4}=0\)

\(\Leftrightarrow\left(x+8\right).\left(\frac{1}{7}+\frac{1}{6}-\frac{1}{5}-\frac{1}{6}\right)=0\)

\(\Leftrightarrow x+8=0\)  ( do \(\left(\frac{1}{7}+\frac{1}{6}-\frac{1}{5}-\frac{1}{6}\right)\ne0\))

\(\Leftrightarrow x=-8\)

Vậy x = - 8

Bài này của lp 7 mà _________ Cách trình bày của lp 7

@@ Học tốt 

Chiyuki Fujito

\(\frac{x+1}{7}+1+\frac{x+2}{6}+1=\frac{x+3}{5}+1+\frac{x+4}{4}+1\)

\(\frac{x+8}{7}+\frac{x+8}{6}=\frac{x+8}{5}+\frac{x+4}{4}\)

\(\left(x+8\right).\left(\frac{1}{7}+\frac{1}{6}-\frac{1}{4}-\frac{1}{5}\right)=0\)

\(x+8=0\)

\(x=-8\)

\(4-3x=10\)

\(3x=-6\)

\(x=-3\)

\(2x-\left(6+4x\right)=8\)

\(2x-6-4x=8\)

\(-2x=8+6\)

\(-2x=14\)

\(x=-7\)

4-3x=10

=> 3x = 4 - 10

=> 3x = -6

=> x = -6 : 3

=> x = -2

\(x\sqrt{x}-3x+4\sqrt{x}-2=x\sqrt{x}-x-2x+2\sqrt{x}+2\sqrt{x}-2\)

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

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

\(ĐKXĐ:x\ne\pm2\)

\(\frac{x}{x+2}+\frac{6}{2-x}=\frac{3x-12}{x^2-4}\)

\(\Leftrightarrow\frac{x}{x+2}-\frac{6}{x-2}-\frac{3x-12}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{x\left(x-2\right)-6\left(x+2\right)-\left(3x-12\right)}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow x^2-2x-6x-12-3x+12=0\)

\(\Leftrightarrow x^2-11x=0\)

\(\Leftrightarrow x\left(x-11\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-11=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=11\end{cases}}\)(tm)

Vậy tập nghiệm của phương trình là \(S=\left\{0;11\right\}\)

\(ĐKXĐ:x\ne\pm2\)

\(\frac{x}{x+2}+\frac{6}{2-x}=\frac{3x-12}{x^2-4}\)

\(\Leftrightarrow\frac{x}{x+2}+\frac{-6}{x-2}-\frac{3x-12}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{-6\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3x-12}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{x\left(x-2\right)-6\left(x+2\right)-\left(3x-12\right)}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow x\left(x-2\right)-6\left(x+2\right)-\left(3x-12\right)=0\)

\(\Leftrightarrow x^2-2x-6x-12-3x+12=0\)

\(\Leftrightarrow x^2-11x=0\)\(\Leftrightarrow x\left(x-11\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-11=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=11\end{cases}}\)( thoả mãn \(ĐKXĐ\))

Vậy tập nghiệm của phương trình là \(S=\left\{0;11\right\}\)

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

\(\Leftrightarrow4x^2+12x+9-3\left(x^2-16\right)=x^2-4x+4+1\)

\(\Leftrightarrow4x^2+12x+9-3x^2+48=x^2-4x+5\)

\(\Leftrightarrow x^2+12x+57=x^2-4x+5\)

\(\Leftrightarrow16x+52=0\)

\(\Leftrightarrow x=-\frac{13}{4}\)

b) \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)

\(\Leftrightarrow\)Xem lại đề !

c) \(x\left(x-1\right)-\left(x-3\right)\left(x+4\right)=5x\)

\(\Leftrightarrow x^2-x-x^2-x+12=5x\)

\(\Leftrightarrow-2x+12=5x\)

\(\Leftrightarrow7x-12=0\)

\(\Leftrightarrow x=\frac{12}{7}\)

d) \(\left(2x+1\right)\left(2x-1\right)=4x\left(x-7\right)-3x\)

\(\Leftrightarrow4x^2-1=4x^2-28x-3x\)

\(\Leftrightarrow28x+3x-1=0\)

\(\Leftrightarrow31x-1=0\)

\(\Leftrightarrow x=\frac{1}{31}\)

a) (2x + 3)² - 3 (x - 4) (x + 4)= (x - 2)² + 1

<=> 4x^2 + 12x + 9 - 3(x^2 - 16) = x^2 - 4x + 4 + 1 

<=> 4x^2 + 12x + 9 - 3x^2 + 48 = x^2 - 4x + 5

<=> x^2 + 12x + 57 = x^2 - 4x + 5

<=> x^2 - x^2 + 12x + 4x + 57 - 5 = 0

<=> 16x + 52 = 0

<=> 16x = -52

<=> x = -13/4

 Đổi : 1h 20 '= \(\frac{4}{3}\) h 
Gọi vận tốc riêng của ca nô la \(a\) ( km/h )( \(a>0\) ) 
\(\Rightarrow\) Vận tốc của ca nô khi xuôi dòng là : \(a+3\)( km/h ) 
Và vận tốc của ca nô khi ngược dòng là :\(a-3\)( km/h ) 
 Ta có phương trình sau : 
\(\frac{4}{3}.\left(a+3\right)=2\left(a-3\right)\)
\(\Leftrightarrow4a+12=6a-18\)
\(\Leftrightarrow a=15\)
Vậy vận tốc riêng của ca nô là : \(15\)km/h

em ko biết vì mới học lớp 5

A)\(ĐKXĐ:x\ne1;2;3;4;5\)

B)Ta có:\(P=\frac{1}{x^2-x}+\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}\)

\(=\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x^2-x\right)-\left(2x-2\right)}+\frac{1}{\left(x^2-2x\right)-\left(3x-6\right)}+\frac{1}{\left(x^2-3x\right)-\left(4x-12\right)}+\frac{1}{\left(x^2-4x\right)-\left(5x-20\right)}\)

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

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

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

nhầm

\(\frac{1}{\left(x-1\right)x}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-3\right)\left(x-2\right)}+\frac{1}{\left(x-4\right)\left(x-3\right)}+\frac{1}{\left(x-5\right)\left(x-4\right)}\)

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

Xin lỗi nha