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

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

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

\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0

\(x-1=0\)

\(x=1\)

Theo đề bài ta có: \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}-\frac{x-4}{5}-\frac{x-5}{6}>0\)

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

<=> \(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}>1\)

<=>\(\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)>1\)

<=> \(\left(x+1\right)\cdot\frac{43}{60}>1\)

<=>\(x+1>\frac{60}{43}\)

<=> x>\(\frac{17}{43}\)

Vậy x>17/43

ĐKXĐ: bạn tự tính nhé

PT tương đương: \(\frac{5}{x-1}-\frac{5}{x-3}=\frac{2}{x+1}-\frac{2}{x-4}\)

<=>\(\frac{5x-15}{\left(x-1\right)\left(x-3\right)}-\frac{5x-5}{\left(x-1\right)\left(x-3\right)}=\frac{2x-8}{\left(x+1\right)\left(x-4\right)}-\frac{2x+2}{\left(x+1\right)\left(x-4\right)}\)

<=>\(\frac{-10}{\left(x-1\right)\left(x-3\right)}=\frac{-10}{\left(x+1\right)\left(x-4\right)}\)

<=>\(\frac{1}{\left(x-1\right)\left(x-3\right)}=\frac{1}{\left(x+1\right)\left(x-4\right)}\)

<=>\(\frac{\left(x+1\right)\left(x-4\right)}{\left(x-1\right)\left(x-3\right)\left(x+1\right)\left(x-4\right)}=\frac{\left(x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)\left(x+1\right)\left(x-4\right)}\)

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

Còn lại bạn từ làm nhé:)

1)

a) \(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}< =>\frac{2\left(x+5\right)}{2\left(3x-6\right)}-\frac{3x-6}{2\left(3x-6\right)}=\frac{3\left(2x-3\right)}{3\left(2x-4\right)}.\)

(đk:x khác \(\frac{1}{2}\))

\(\frac{2x+10}{6x-12}-\frac{3x-6}{6x-12}=\frac{6x-9}{6x-12}< =>2x+10-3x+6=6x-9< =>x=\frac{25}{7}\)

Vậy x=\(\frac{25}{7}\)

b) /7-2x/=x-3 \(x\ge\frac{7}{2}\)

(đk \(x\ge3,\frac{7}{2}< =>x\ge\frac{7}{2}\))

\(\Rightarrow\orbr{\begin{cases}7-2x=x-3\\7-2x=-\left(x-3\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\left(< \frac{7}{2}\Rightarrow l\right)\\x=4\left(tm\right)\end{cases}}}\)

Vậy x=4

2)

\(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}>\frac{x-4}{5}+\frac{x-5}{6}\)

\(\Leftrightarrow\frac{30\left(x-1\right)}{60}+\frac{20\left(x-2\right)}{60}+\frac{15\left(x-3\right)}{60}-\frac{12\left(x-4\right)}{60}-\frac{10\left(x-5\right)}{60}>0\)

\(\Leftrightarrow30x-30+20x-40+15x-45-12x+48-10x+50>0\Leftrightarrow43x-17>0\Leftrightarrow x>\frac{17}{43}\)

\(\text{ĐKXĐ : }x\notin\left\{0;-1;-2;-3\right\}\). Ta biến đổi phương trình như sau :

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

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

\(\Leftrightarrow\frac{5+x}{x}+\frac{5+x}{x+3}=\frac{5+x}{x+1}+\frac{5+x}{x+2}\)

\(\Leftrightarrow(5+x)\left(\frac{1}{x}+\frac{1}{x+3}-\frac{1}{x+1}-\frac{1}{x+2}\right)=0\)

\(\Leftrightarrow5+x=0\text{ (1) hoặc }\frac{1}{x}+\frac{1}{x+3}-\frac{1}{x+1}-\frac{1}{x+2}=0\text{ (2) }\).

Ta có :

\(\left(1\right)\Leftrightarrow x=-5\);

\(\left(2\right)\Leftrightarrow\frac{1}{x}+\frac{1}{x+3}=\frac{1}{x+1}+\frac{1}{x+2}\Leftrightarrow\frac{2x+3}{x\left(x+3\right)}=\frac{2x+3}{\left(x+1\right)\left(x+2\right)}\)

\(\Leftrightarrow\left(2x+3\right)\left(\frac{1}{x^2+3x}-\frac{1}{x^2+3x+2}\right)=0\)

\(\Leftrightarrow2x+3=0\text{ hoặc }\frac{1}{x^2+3x}-\frac{1}{x^2+3x+2}=0\).

Tóm lại, phương trình đã cho có tập nghiệm \(S=\left\{-5;-\frac{3}{2}\right\}\).

A . 3x + 2(x + 1) = 6x - 7

<=> 3x + 2x + 2 = 6x -7

<=> 5x - 6x = -7 - 2

<=> -x = -9

<=> x =9

B . \(\frac{x+3}{5}\).< \(\frac{5-x}{3}\)

=> 3(x +3) < 5(5 -x)

<=> 3x+9 < 25 - 5x

<=> 3x + 5x < 25 - 9

<=> 8x < 16

<=> x < 2

C . \(\frac{5}{x+1}\)+ \(\frac{2x}{x^2-3x-4}\)=\(\frac{2}{x-4}\)

<=> \(\frac{5}{x+1}\)+ \(\frac{2x}{x^2+x-4x-4_{ }}\)= \(\frac{2}{x-4}\)

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

<=> 5(x - 4) + 2x = 2(x +1)

<=> 5x - 20 + 2x = 2x + 2

<=>7x - 2x = 2 + 20

<=> 5x = 22

<=> x =\(\frac{22}{5}\)

cho tam giác abc vuông tại a và đường cao ah =12cm, ch = 5cm. tính sin b sin c

ai giải giúp mình bài toán này với mk đang cần rất gấp

\(\text{a) }\frac{6}{x-4}-\frac{x}{x+2}=\frac{6}{x-4}.\frac{x}{x+2}\)

\(ĐKXĐ:x\ne-2;x\ne4\)

\(MTC:\left(x-4\right)\left(x+2\right)\)

\(\Leftrightarrow\frac{6\left(x+2\right)}{\left(x-4\right)\left(x+2\right)}-\frac{x\left(x-4\right)}{\left(x-4\right)\left(x+2\right)}=\frac{6x}{\left(x-4\right)\left(x+2\right)}\)

\(\Rightarrow6\left(x+2\right)-x\left(x-4\right)=6x\)

\(\Leftrightarrow6x+12-x^2+4x=6x\)

\(\Leftrightarrow6x+12-x^2+4x-6x=0\)

\(\Leftrightarrow-x^2+4x+12=0\)

\(\Leftrightarrow-\left(x^2-4x-12\right)=0\)

\(\Leftrightarrow x^2-4x-12=0\)

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

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

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

\(\Leftrightarrow x=-2\left(\text{loại}\right)\text{ hoặc }x=6\left(\text{nhận}\right)\)

Vậy \(S=\left\{6\right\}\)

\(\text{b) }\frac{2x+3}{2x-1}=\frac{x-3}{x+5}\)

\(ĐKXĐ:x\ne\frac{1}{2};x\ne-5\)

\(\Leftrightarrow\left(2x+3\right)\left(x+5\right)=\left(2x-1\right)\left(x-3\right)\left[\text{Tỉ lệ thức}\right]\)

\(\Leftrightarrow2x^2+10x+3x+15=2x^2-6x-x+3\)

\(\Leftrightarrow2x^2+13x+15=2x^2-7x+3\)

\(\Leftrightarrow2x^2+13x-2x^2+7x=3-15\)

\(\Leftrightarrow20x=-12\)

\(\Leftrightarrow x=\frac{-12}{20}=\frac{-3}{5}\)

Vậy \(S=\left\{\frac{-3}{5}\right\}\)

cộng mỗi vế với 2, mõi phân số cộng với 1 được:
 (x+5)/x +(x+5)/X+1=(x+5)/x+2  + (x+5)/x+3 
đổi vế, đặt x+5 ra ngoài, bên trong ngoặc là 1 số khác 0
nên x+5=0 =>x=-5
ĐKXĐ: x khác 0;-1;-2;-3