Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
<=> 1 - x + 3(x + 1) = 2x + 3
<=> 1 - x + 3x + 3 = 2x + 3
<=> 1 - x + 3x + 3 - 2x = 3
<=> 4 = 3 (vô lý)
=> pt vô nghiệm
b) ĐKXĐ: \(x\ne1;x\ne2\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
<=> (x - 2)(2 - x) - 5(x + 1)(2 - x) = 15(x - 2)
<=> 2x - x2 - 4 + 2x - 5x - 5x2 + 10 = 15x - 30
<=> -x + 4x2 - 14 = 15x - 30
<=> x - 4x2 + 14 = 15x - 30
<=> x - 4x2 + 14 + 15x - 30 = 0
<=> 16x - 4x2 - 16 = 0
<=> 4(4x - x2 - 4) = 0
<=> -x2 + 4x - 4 = 0
<=> x2 - 4x + 4 = 0
<=> (x - 2)2 = 0
<=> x - 2 = 0
<=> x = 2 (ktm)
=> pt vô nghiệm
c) xem bài 4 ở đây: Câu hỏi của gjfkm
d) ĐKXĐ: \(x\ne1;x\ne2;x\ne3\)
\(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
<=> \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}+\frac{x+1}{\left(x-1\right)\left(x-3\right)}=\frac{2x+5}{\left(x-1\right)\left(x-3\right)}\)
<=> (x + 4)(x - 3) + (x + 1)(x - 2) = (2x + 5)(x - 2)
<=> x2 - 3x + 4x - 12 + x2 - 2x + x - 2 = 2x2 - 4x + 5x - 10
<=> 2x2 - 14 = 2x2 + x - 10
<=> 2x2 - 14 - 2x2 = x - 10
<=> -14 = x - 10
<=> -14 + 10 = x
<=> -4 = x
<=> x = -4
d, (x2 + 4x + 8)2 + 3x(x2 + 4x + 8) + 2x2 = 0
Đặt x2 + 4x + 8 = t ta được:
t2 + 3xt + 2x2 = 0
\(\Leftrightarrow\) t2 + xt + 2xt + 2x2 = 0
\(\Leftrightarrow\) t(t + x) + 2x(t + x) = 0
\(\Leftrightarrow\) (t + x)(t + 2x) = 0
Thay t = x2 + 4x + 8 ta được:
(x2 + 4x + 8 + x)(x2 + 4x + 8 + 2x) = 0
\(\Leftrightarrow\) (x2 + 5x + 8)[x(x + 4) + 2(x + 4)] = 0
\(\Leftrightarrow\) (x2 + 5x + \(\frac{25}{4}\) + \(\frac{7}{4}\))(x + 4)(x + 2) = 0
\(\Leftrightarrow\) [(x + \(\frac{5}{2}\))2 + \(\frac{7}{4}\)](x + 4)(x + 2) = 0
Vì (x + \(\frac{5}{2}\))2 + \(\frac{7}{4}\) > 0 với mọi x
\(\Rightarrow\left[{}\begin{matrix}x+4=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-2\end{matrix}\right.\)
Vậy S = {-4; -2}
Mình giúp bn phần khó thôi!
Chúc bn học tốt!!
c) \(\frac{1}{x-1}\)+\(\frac{2x^2-5}{x^3-1}\)=\(\frac{4}{x^2+x+1}\) (ĐKXĐ:x≠1)
⇔\(\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)+\(\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}\)=\(\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
⇒x2+x+1+2x2-5=4x-4
⇔3x2-3x=0
⇔3x(x-1)=0
⇔x=0 (TMĐK) hoặc x=1 (loại)
Vậy tập nghiệm của phương trình đã cho là:S={0}
\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x^2+x}\)
b, \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{y^2-xy-xy+x^2}{\left(xy-x^2\right)\left(y^2-xy\right)}=\frac{x^2+y^2}{xy^3-xyxy-xyxy+x^3y}\)Tu rut gon tiep
c, tt
d, cx r
a) \(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}\)
\(=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)
b) \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)
\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)
c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)
\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)
Làm ngắn gọn thôi nhé :v
\(A=\frac{2x}{x^2-3x}+\frac{2x}{x^2-4x+3}+\frac{x}{x-1}\)
\(A=\frac{x^5-3x^4-3x^3+11x^2-6x}{x^5-8x^2+22x^2-24x+9}\)
\(A=\frac{x^4-3x^3-3x^2+11x-6}{x^4-8x^3+22x^2-24x+9}\)
\(A=\frac{\left(x-1\right)\left(x-1\right)\left(x+2\right)\left(x-3\right)}{\left(x-1\right)\left(x-1\right)\left(x-3\right)\left(x-3\right)}\)
\(A=\frac{x+2}{x-3}\)
\(B=\frac{x}{x+2}+\frac{2}{x-2}-\frac{4x}{4-x^2}\)
\(B=\frac{-x^4-4x^3+16x+16}{-x^4+8x^2-16}\)
\(B=\frac{\left(-x-2\right)\left(x+2\right)\left(x+2\right)\left(x-2\right)}{\left(-x-2\right)\left(x-2\right)\left(x+2\right)\left(x-2\right)}\)
\(B=\frac{x+2}{x-2}\)
\(C=\frac{1+x}{3-x}-\frac{1-2x}{3+x}-\frac{x\left(1-x\right)}{9-x^2}\)
\(C=\frac{1+x}{3-x}-\left(\frac{1-2x}{3+x}\right)-\frac{x\left(1-x\right)}{9-x^2}\)
\(C=\frac{10x}{-x^2+9}\)
\(D=\frac{5}{2x^2+6x}-\frac{4-3x^2}{x^2-9}-3\)
\(D=\frac{5}{2x^2+6x}-\left(\frac{4-3x^2}{x^2-9}\right)-3\)
\(D=\frac{51x^2+138x-45}{2x^4+6x^2-18x^2-54x}\)
\(D=\frac{3\left(17x-5\right)\left(x+3\right)}{2x\left(x+3\right)\left(x+3\right)\left(x-2\right)}\)
\(D=\frac{51x-15}{2x^3-18x}\)
\(E=\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\frac{3x-2}{x^2+2x+1}\)
\(E=\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\left(\frac{3x-2}{x^2+2x+1}\right)\)
\(E=\frac{10x^4-10}{x^6-3x^4+3x^2-1}\)
\(E=\frac{10\left(x^2+1\right)\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x+1\right)\left(x+1\right)\left(x-1\right)\left(x-1\right)\left(x-1\right)}\)
\(E=\frac{10x^2+10}{x^4-2x+1}\)
Bài 1:
d)ĐKXĐ: \(x\ne8\)
Ta có: \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{13x-102}{3x-24}\)
\(\Leftrightarrow\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3x-24}=0\)
\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3\left(x-8\right)}=0\)
MTC=24(x-8)
\(\Leftrightarrow\frac{36}{24\left(x-8\right)}+\frac{72x-480}{24\left(x-8\right)}+\frac{3x-24}{24\left(x-8\right)}-\frac{104x-816}{24\left(x-8\right)}=0\)
\(\Leftrightarrow36+72x-480+3x-24-104x+816=0\)
\(\Leftrightarrow348-29x=0\)
\(\Leftrightarrow-29x+348=0\)
\(\Leftrightarrow x=\frac{-348}{-29}=12\)
Vậy: x=12
e) ĐKXĐ: \(x\ne\pm1\)
Ta có: \(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)
\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4x+4}+\frac{12x-1}{4-4x}=0\)
\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4\left(x+1\right)}+\frac{12x-1}{4\left(1-x\right)}=0\)
MTC=4(x+1)(x-1)
\(\Leftrightarrow\frac{24}{4\left(x-1\right)\left(x+1\right)}+\frac{20x^2-20}{4\left(x-1\right)\left(x+1\right)}-\frac{8x^2-9x+1}{4\left(x-1\right)\left(x+1\right)}-\frac{12x^2-11x-1}{4\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow24+20x^2-20-8x^2+9x-1-12x^2+11x+1=0\)
\(\Leftrightarrow20x+4=0\)
\(\Leftrightarrow20x=-4\)
\(\Leftrightarrow x=-\frac{4}{20}=-0,2\)(loại)
Vậy: x không có giá trị
g) Ta có: \(\frac{\frac{x+1}{x-1}-\frac{x-1}{x+1}}{1+\frac{x+1}{x-1}}=\frac{1}{2}\)
\(\Leftrightarrow\frac{\frac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}}{\frac{x-1}{x-1}+\frac{x+1}{x-1}}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}}{\frac{2x}{x-1}}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x-1}{2x}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{4x\cdot\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\cdot2x}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{2}=0\)
MTC=2(x+1)
\(\Leftrightarrow\frac{2}{2\left(x+1\right)}-\frac{x+1}{2\left(x+1\right)}=0\)
\(\Leftrightarrow2-x+1=0\)
\(\Leftrightarrow1-x=0\)
\(\Leftrightarrow x=1\)(loại vì không thỏa mãn ĐKXĐ)
Vậy: x không có giá trị
a)\(\frac{3x-2}{5}\ge\frac{x}{2}+0,8\) va \(1-\frac{2x-5}{6}>\frac{3-x}{4}\)
\(\cdot\frac{3x-2}{5}\ge\frac{x}{2}+0,8\)
\(=\frac{2\left(3x-2\right)}{10}\ge\frac{5x}{10}+\frac{8}{10}\)
\(\Rightarrow2\left(3x-2\right)\ge5x+8\)
\(=6x-4\ge5x+8\)
\(=6x-5x\ge8+4\)
\(x\ge12\)(1)
\(\cdot1-\frac{2x-5}{6}>\frac{3-x}{4}\)
\(=\frac{12}{12}-\frac{2\left(2x-5\right)}{12}>\frac{3\left(3-x\right)}{12}\)
\(\Rightarrow12-2\left(2x-5\right)>3\left(3-x\right)\)
\(=12-4x+10>9-3x\)
\(=-4x+3x>9-12-10\)
\(=-x>-13\)
\(=x< 13\) (2)
Từ (1) và (2) => \(13>x\ge12\)=> x=12
a) 4 ( x + 5 )( x + 6 )( x + 10 )( x + 12 ) = 3x2
Do x = 0 không là nghiệm pt nên chia 2 vế pt cho \(x^2\ne0\), ta được :
\(\frac{4}{x^2}\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3\)
\(\Leftrightarrow4\left(x+\frac{60}{x}+17\right)\left(x+\frac{60}{x}+16\right)=3\)
Đến đây ta đặt \(x+\frac{60}{x}+16=t\left(1\right)\)
Ta được :
\(4t\left(t+1\right)=3\Leftrightarrow4t^2+4t-3=0\Leftrightarrow\left(2t+3\right)\left(2t-1\right)=0\)
Từ đó ta lắp vào ( 1 ) tính được x
bạn không ghi yêu cầu nên mình làm như này
1) \(\frac{1}{x-3}\) và \(\frac{5}{x^2-3x}\)
Ta có: \(1.\left(x^2-3x\right)=x^2-3x\)
\(\left(x-3\right).5=5x-15\)
\(\Rightarrow x^2-3x\ne5x-15\)
\(\Rightarrow1.\left(x^2-3x\right)\ne\left(x-3\right).5\)
Vậy: \(\frac{1}{x-3}\ne\frac{5}{x^2-3x}\)
2) \(\frac{x}{x^2+x}\) và \(\frac{2}{x-1}\) và \(\frac{x+2}{x^2-1}\)
Ta có: \(x.\left(x-1\right)=x^2-x\)
\(2.\left(x^2+x\right)=2x^2+2x\)
\(\Rightarrow x^2-x\ne2x^2+2x\)
\(\Rightarrow x.\left(x-1\right)\ne2.\left(x^2+x\right)\)
\(\Rightarrow\frac{1-3x}{2x}\ne\frac{2}{x-1}\) (1)
Ta lại có: \(2.\left(x^2-1\right)=2x^2-2\)
\(\left(x-1\right)\left(x+2\right)=x^2+2x-x-2\)
\(=x^2-x-2\)
\(\Rightarrow2x^2-2\ne x^2-x-2\)
\(\Rightarrow2.\left(x^2-1\right)\ne\left(x-1\right)\left(x+2\right)\)
\(\Rightarrow\frac{2}{x-1}\ne\frac{x+2}{x^2-1}\) (2)
Từ (1) và (2) => \(\frac{x}{x^2+x}\ne\frac{2}{x-1}\ne\frac{x+2}{x^2-1}\)
3) \(\frac{1-3x}{2x}\) và \(\frac{3x-2}{2x-1}\) và \(\frac{3x-2}{4x^2-2x}\)
Ta có:\(\left(1-3x\right)\left(2x-1\right)=2x-1-6x^2+3x\)
\(=5x-1-6x^2\)
\(2x.\left(3x-2\right)=6x^2-4x\)
\(\Rightarrow5x-1-6x^2\ne6x^2-4x\)
\(\Rightarrow\left(1-3x\right)\left(2x-1\right)\ne2x\left(3x-2\right)\)
\(\Rightarrow\frac{1-3x}{2x}\ne\frac{3x-2}{2x-1}\)(1)
Ta lại có: \(\left(3x-2\right)\left(4x^2-2x\right)=12x^2-6x^2-8x^2+4x\)
\(=12x^3-14x^2+4x\)
\(\left(2x-1\right)\left(3x-2\right)=6x^2-4x-3x+2\)
\(=6x^2-7x+2\)
\(\Rightarrow12x^3-14x^2+4x\ne6x^2-7x+2\)
\(\Rightarrow\left(3x-2\right)\left(4x^2-2x\right)\ne\left(2x-1\right)\left(3x-2\right)\)
\(\Rightarrow\frac{3x-2}{2x-1}\ne\frac{3x-2}{4x^2-2x}\) (2)
Từ (1) và (2) => \(\frac{1-3x}{2x}\ne\frac{3x-2}{2x-1}\ne\frac{3x-2}{4x^2-2x}\)