GPT: \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\)= \(\frac{3}{10}\)
Tìm x, biết:
\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\left(x\notin-2;-5;-10;-17\right)\)
\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)
Với \(x\notin1;3;8;20\)
\(\frac{x+1}{10}+\frac{2+1}{11}\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\frac{x-10}{30}+\frac{x-14}{43}+\frac{x-5}{95}+\frac{x-148}{8}=0\)
Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0
Cho biểu thức
A=\(\left[\frac{3\left(x+2\right)}{2\left(x^3+x^2+x+1\right)}+\frac{2x^2-x-10}{2\left(x^3-x^2+x-1\right)}\right]:\left[\frac{5}{x^2+1}+\frac{3}{2\left(x+1\right)}-\frac{3}{2\left(x+1\right)}\right]\)
a) gpt \(\left(x-1\right)\left(x+2\right)+4\left(x-1\right)\sqrt{\frac{x+2}{x-1}}=12\)
b) ghpt \(\left\{\begin{matrix}2\sqrt{x}\left(1+\frac{1}{x+y}\right)=3\\2\sqrt{y}\left(1-\frac{1}{x+y}\right)=1\end{matrix}\right.\)
a/ \(\left(x-1\right)\left(x+2\right)+4\left(x-1\right)\sqrt{\frac{x+2}{x-1}}=12\)
Điều kiện: \(\left[\begin{matrix}x\le-2\\x>1\end{matrix}\right.\)
Xét \(x\le-2\) thì ta có
\(\left(x-1\right)\left(x+2\right)+4\left(x-1\right)\sqrt{\frac{x+2}{x-1}}=12\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)-4\sqrt{\left(x-1\right)\left(x+2\right)}=12\)
Đặt \(\sqrt{\left(x-1\right)\left(x+2\right)}=a\left(a\ge0\right)\) thì pt thành
\(a^2-4a-12=0\)
\(\Leftrightarrow\left[\begin{matrix}a=-2\left(l\right)\\a=6\end{matrix}\right.\)
\(\Rightarrow\sqrt{\left(x-1\right)\left(x+2\right)}=6\)
\(\Leftrightarrow x^2+x-38=0\)
\(\Leftrightarrow\left[\begin{matrix}x=-\frac{1}{2}+\frac{3\sqrt{17}}{2}\left(l\right)\\x=-\frac{1}{2}-\frac{3\sqrt{17}}{2}\end{matrix}\right.\)
Trường hợp x > 1 làm tương tự nhé
Bài 2 :
a, \(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
b, \(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{\left(x-20\right)}=\frac{-3}{4}\)
Tìm x,biết
a, \(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
Với x ∉ -2,-5,-10,-17
b,\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)
Với x∉1,3,8,20
c,\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
c) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(\Leftrightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right).\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
\(\Leftrightarrow x-2010=0\)
\(\Leftrightarrow x=0+2010\)
\(\Rightarrow x=2010\)
Vậy \(x=2010.\)
Mình chỉ làm câu c) thôi nhé.
Chúc bạn học tốt!
a,\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|=4x\)
b,\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
a) Dễ thấy VT > 0;mà VT=VP
=>VP > 0 => 4x > 0=> x > 0
=>\(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{3}\right|=x+\frac{1}{3};\left|x+\frac{1}{6}\right|=x+\frac{1}{6}\)
=>BT đầu tương đương \(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{6}\right)=4x\)
\(=>3x+1=4x=>x=1\)
a) Để đẳng thức xảy ra thì: x>0 (vì: \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|>0\) )
Khi đó: \(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{3}\right|=x+\frac{1}{3};\left|x+\frac{1}{6}\right|=x+\frac{1}{6}\)
=>\(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{6}=4x\)
<=>x=1
Vậy x=1
b)Điều kiện: \(x\ne-3;-10;-21;-34\)
\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
<=>\(\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
<=>\(\frac{1}{x+3}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
=>x+34-x-3=x
<=>x=31 (nhận)
Vậy x=31
a,\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|=4x\)
Ta có: \(\begin{cases}\left|x+\frac{1}{2}\right|\ge0\\\left|x+\frac{1}{3}\right|\ge0\\\left|x+\frac{1}{6}\right|\ge0\end{cases}\)
\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|\ge0\)
\(\Rightarrow4x\ge0\)
\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|=x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{6}\)
Khi đó, ta có: \(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{6}=4x\)
\(\Rightarrow3x+1=4x\)
\(\Rightarrow x=1\)
b) Từ đề suy ra:
\(\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\Rightarrow\frac{1}{x+3}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\Rightarrow\frac{x+34}{\left(x+3\right)\left(x+34\right)}-\frac{x+3}{\left(x+3\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\Rightarrow\frac{31}{\left(x+3\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\Rightarrow x=31\)
Giải phương trình sau:
a) \(\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{3}{1-x}=\frac{5}{x+5}\)
b) \(\frac{x-1}{x+2}+\frac{x+3}{x-4}=\frac{2}{\left(x-2\right)\left(4-x\right)}\)
c) \(\frac{7x-3}{x-x^3}=\frac{1}{x-1}-\frac{5}{x\left(x-1\right)}\)
d) \(\frac{1}{\left(x+2\right)}+\frac{1}{\left(x+3\right)}=\frac{1}{\left(x+2\right)\left(x+3\right)}\)
Giai phương trình sau :
a) \(\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{3}{1-x}=\frac{5}{x+5}\)
ĐKXĐ : \(x\ne1;x\ne-5\)
Với điều kiện trên ta có :
\(\Leftrightarrow\)\(\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{-3}{x-1}=\frac{5}{x+5}\)
\(\Leftrightarrow10-3\left(x+5\right)=5\left(x-1\right)\)
\(\Leftrightarrow10-3x-15=5x-5\)
\(\Leftrightarrow-8x=0\)
\(\Leftrightarrow x=0\) (nhận)
Vậy : \(S=\left\{0\right\}\)
d) \(\frac{1}{\left(x+2\right)}+\frac{1}{\left(x+3\right)}=\frac{1}{\left(x+2\right)\left(x+3\right)}\)
ĐKXĐ : \(x\ne-2;x\ne-3\)
\(\Leftrightarrow x+3+x+2=1\)
\(\Leftrightarrow2x=-4\)
\(\Leftrightarrow x=-2\) (không nhận)
Vậy : \(S=\varnothing\)
GPT:
\(\left(1\right)\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)
\(\left(2\right)\frac{x-3}{x+3}-\frac{x+3}{x-3}=-\frac{5}{x^2-9}\)
\(\left(1\right)\Leftrightarrow2x-3x^2+11-33x=6x-4-15x^2+10x\)
\(\Leftrightarrow12x^2-47x+15=0\)
\(\Delta=47^2-4.12.15=1489,\sqrt{\Delta}=\sqrt{1489}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{47+\sqrt{1489}}{24}\\x=\frac{47-\sqrt{1489}}{24}\end{cases}}\)
\(\left(2\right)\Leftrightarrow\frac{\left(x-3\right)^2-\left(x+3\right)^2}{x^2-9}=\frac{-5}{x^2-9}\)
\(\Leftrightarrow\left(x-3\right)^2-\left(x+3\right)^2=-5\)
\(\Leftrightarrow x^2-6x+9-x^2-6x-9=-5\)
\(\Leftrightarrow-12x=-5\Leftrightarrow x=\frac{5}{12}\)
(2-3x)(x+11)=(3x-2)(2-5x)
<=>(3x-2)(2-5x)-(2-3x)(x+11)=0
<=>(3x-2)(2-5x)+(3x-2)(x+11)=0
<=>(3x-2)[2-5x+x+11]=0
<=>(3x-2)(13-4x)=0
<=>\(\orbr{\begin{cases}3x-2=0\\13-4x=0\end{cases}}\)
<=>\(\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{13}{4}\end{cases}}\)
\(\frac{x-3}{x+3}-\frac{x+3}{x-3}=-\frac{5}{x^2-9}\)
Đk:\(x\ne-3;x\ne3\)(*)
Với đk trên pt tương đương với:
\(\frac{\left(x-3\right)^2-\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}=-\frac{5}{\left(x+3\right)\left(x-3\right)}\)
\(x^2-6x+9-x^2-6x-9=-5.-12x=-5\)
\(x=\frac{15}{12}\left(tmđk\right)\)(*)
Giải phương trình:
1. \(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
2. \(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
3. \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
4. \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
5. \(\frac{x-4}{5}-\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+2}{6}\)
6. \(\frac{\left(x+2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)
7. \(\frac{\left(x+2\right)^2}{8}-2\left(2x-1\right)=25+\frac{\left(x-2\right)^2}{8}\)
8.\(\frac{7x^2-14x-5}{5}=\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}\)
9. \(\frac{\left(2x-3\right)\left(2x+3\right)}{8}=\frac{\left(x-4\right)^2}{6}+\frac{\left(x-2\right)^2}{3}\)
10. \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)