Giải phương trình
a. \(\frac{3}{x-1}\)= \(\frac{3x+2}{1-x^2}\)- \(\frac{4}{x+1}\)
b. x2 - 9 = (x+3)(1-x)
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\(a\text{) }7-\left(2x+4\right)=-\left(x+4\right)\)
\(\Leftrightarrow7-2x-4=-x-4\)
\(\Leftrightarrow x=7\)
\(b\text{) }\frac{3x-1}{3}=\frac{2-x}{2}\)
\(\Leftrightarrow2\left(3x-1\right)=3\left(2-x\right)\)
\(\Leftrightarrow6x-2=6-3x\)
\(\Leftrightarrow9x=8\Leftrightarrow x=\frac{8}{9}\)
\(c\text{) }\frac{2\left(3x+5\right)}{3}-\frac{x}{2}=5-\frac{3\left(x+1\right)}{4}\)
\(\Leftrightarrow8\left(3x+5\right)-6x=60-9\left(x+1\right)\)
\(\Leftrightarrow24x+40-6x=60-9x-9\)
\(\Leftrightarrow27x=11\Leftrightarrow x=\frac{11}{27}\)
\(d\text{) }x^2-4x+4=9\)
\(\Leftrightarrow\left(x-2\right)^2=3^2\)
\(\Leftrightarrow x-2=3\Leftrightarrow x=5\)
\(e\text{) }\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{5x-8}{x^2-4}\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)-x\left(x+2\right)=5x-8\)
\(\Leftrightarrow x^2-x-2x+3-x^2-2x=5x-8\)
\(\Leftrightarrow11-10x=0\Leftrightarrow x=\frac{11}{10}\)
a: \(\Leftrightarrow x^2-4-4x^2-4x-1-2x+3x^2=0\)
=>-6x-5=0
=>-6x=5
hay x=-5/6
b: \(\Leftrightarrow2x^3+8x^2+8x-8x^2-2x^3+16=0\)
=>8x+16=0
hay x=-2
c: \(\Leftrightarrow x^3-6x^2+12x-8+9x^2-1-x^3-3x^2-3x-1=0\)
=>9x-10=0
hay x=10/9
d: \(\Leftrightarrow10x-15-20x+28=19-2x^2-4x-2\)
\(\Leftrightarrow-10x+13+2x^2+4x-17=0\)
\(\Leftrightarrow2x^2-6x-4=0\)
\(\Leftrightarrow x^2-3x-2=0\)
\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-2\right)=9+8=17>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{17}}{2}\\x_2=\dfrac{3+\sqrt{17}}{2}\end{matrix}\right.\)
Bài 1:
1, \(\frac{2x-5}{x+5}=3\) (ĐKXĐ: x \(\ne\) -5)
\(\Leftrightarrow\) \(\frac{2x-5}{x+5}=\frac{3\left(x+5\right)}{x+5}\)
\(\Rightarrow\) 2x - 5 = 3(x + 5)
\(\Leftrightarrow\) 2x - 5 = 3x + 15
\(\Leftrightarrow\) 2x - 3x = 15 + 5
\(\Leftrightarrow\) -x = 20
\(\Leftrightarrow\) x = -20 (TMĐKXĐ)
Vậy S = {-20}
2, \(\frac{4}{x+1}=\frac{3}{x-2}\) (ĐKXĐ: x \(\ne\) -1; x \(\ne\) 2)
\(\Leftrightarrow\) \(\frac{4\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow\) 4(x - 2) = 3(x + 1)
\(\Leftrightarrow\) 4x - 8 = 3x + 3
\(\Leftrightarrow\) 4x - 3x = 3 + 8
\(\Leftrightarrow\) x = 11 (TMĐKXĐ)
Vậy S = {11}
3, \(\frac{5}{2x-3}=\frac{1}{x-4}\) (ĐKXĐ: x \(\ne\) \(\frac{3}{2}\); x \(\ne\) 4)
\(\Leftrightarrow\) \(\frac{5\left(x-4\right)}{\left(2x-3\right)\left(x-4\right)}=\frac{2x-3}{\left(2x-3\right)\left(x-4\right)}\)
\(\Rightarrow\) 5(x - 4) = 2x - 3
\(\Leftrightarrow\) 5x - 20 = 2x - 3
\(\Leftrightarrow\) 5x - 2x = -3 + 20
\(\Leftrightarrow\) 3x = 17
\(\Leftrightarrow\) x = \(\frac{17}{3}\) (TMĐKXĐ)
Vậy S = {\(\frac{17}{3}\)}
Bài 2:
1, \(\frac{1}{x-1}+\frac{2}{x+1}=\frac{5x-3}{x^2-1}\) (ĐKXĐ: x \(\ne\) \(\pm\) 1)
\(\Leftrightarrow\) \(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{5x-3}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow\) x + 1 + 2(x - 1) = 5x - 3
\(\Leftrightarrow\) x + 1 + 2x - 2 = 5x - 3
\(\Leftrightarrow\) 3x - 1 = 5x - 3
\(\Leftrightarrow\) 3x - 5x = -3 + 1
\(\Leftrightarrow\) -2x = -2
\(\Leftrightarrow\) x = 1 (KTM)
\(\Rightarrow\) Pt vô nghiệm
Vậy S = \(\varnothing\)
2, \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\) (ĐKXĐ: x \(\ne\) 2; x \(\ne\) 0)
\(\Leftrightarrow\) \(\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
\(\Rightarrow\) x(x + 2) - x + 2 = 2
\(\Leftrightarrow\) x2 + 2x - x + 2 = 2
\(\Leftrightarrow\) x2 + x = 2 - 2
\(\Leftrightarrow\) x2 + x = 0
\(\Leftrightarrow\) x(x + 1) = 0
\(\Leftrightarrow\) x = 0 hoặc x + 1 = 0
\(\Leftrightarrow\) x = 0 và x = -1
Ta có: x = 0 KTM đkxđ
\(\Rightarrow\) x = -1
Vậy S = {-1}
3, \(\frac{5}{x-3}-\frac{3}{x+3}=\frac{3x}{x^2-9}\) (ĐKXĐ: x \(\ne\) \(\pm\) 3)
\(\Leftrightarrow\) \(\frac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{3x}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow\) 5(x + 3) - 3(x - 3) = 3x
\(\Leftrightarrow\) 5x + 15 - 3x + 9 = 3x
\(\Leftrightarrow\) 2x + 24 = 3x
\(\Leftrightarrow\) 2x - 3x = 24
\(\Leftrightarrow\) -x = 24
\(\Leftrightarrow\) x = -24 (TMĐKXĐ)
Vậy S = {-24}
Chúc bn học tốt!!
Mình tính mãi vẫn có chỗ sai, mong bạn thông cảm!!
Mình bt mình sai đâu r Garuda
câu 3 bài 3 cuối có cái đoạn 2x + 24 = 3x
\(\Leftrightarrow\) 2x - 3x = -24 (đoạn kia mình ghi là 24 nên quên không đổi dấu)
\(\Leftrightarrow\) -x = -24
\(\Leftrightarrow\) x = 24
Vậy S = {24}
(mình sửa lại rồi nha, chắc hết chỗ sai rồi)
\(\frac{x-3}{5}-\frac{2x-1}{10}=\frac{x+1}{2}+\frac{1}{4}\)
\(< =>\frac{\left(x-3\right).4}{20}-\frac{\left(2x-1\right).2}{20}=\frac{\left(x+1\right).10}{20}+\frac{5}{20}\)
\(< =>4x-12-4x+2=10x+10+5\)
\(< =>10x=-10-10-5=-25\)
\(< =>x=-\frac{25}{10}=-\frac{5}{2}\)
\(\frac{x+3}{2}-\frac{2x-1}{3}-1=\frac{x+5}{5}\)
\(< =>\frac{\left(x+3\right).15}{30}-\frac{\left(2x-1\right).10}{30}-\frac{30}{30}=\frac{\left(x+5\right).5}{30}\)\(< =>15x+45-20x+10-30=5x+25\)
\(< =>-5x+25=5x+25< =>10x=0< =>x=0\)
8,
b, (-x2+12x+4)/(x2+3x-4) = 12/(x+4) + 12/(3x-3)
(=) (-x2+12x+4)/(x-1)(x+4) -12(x-1)/(x-1)(x+4) - 4(x+4)/(x-1)(x+4) = 0
(=) -x2 +12x +4 -12x +12 -4x -16 = 0
(=) -x2 -4x = 0
(=) -x(x+4) = 0
(=) -x = 0 hoặc x +4 = 0
(=) x=0 hoặc x=-4
Vậy S={0;4}
Chúc bạn học tốt.
Ta thấy \(\left(x-3\right)\left(2x+3\right)=2x^2-3x-9.\)
\(\left(1\right)\Leftrightarrow\frac{x}{x-3}-\frac{2x^2+9}{\left(x-3\right)\left(2x+3\right)}=\frac{1}{2x+3}\)
ĐK: \(x\ne3\)và \(x\ne-\frac{3}{2}\)
\(\Rightarrow x\left(2x+3\right)-2x^2-9=x-3\)
\(\Leftrightarrow2x^2+3x-2x^2-9=x-3\Leftrightarrow2x=6\Leftrightarrow x=2\)
Thỏa mãn ĐK
Các trường hợp khác làm tương tự
a/ ĐK: \(x\ne\pm3\)
\(\frac{2x-3}{x^2-9}+\frac{2x\left(x+3\right)}{x^2-9}+\frac{5\left(x-3\right)}{x^2-9}=0\)
\(\Leftrightarrow2x-3+2x^2+6x+5x-15=0\)
\(\Leftrightarrow2x^2+13x-18=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{-13+\sqrt{313}}{4}\\x=\frac{-13-\sqrt{313}}{4}\end{matrix}\right.\)
b/ ĐKXĐ: \(x\ne1;-3\)
\(\frac{\left(3x-1\right)\left(x+3\right)}{x^2+2x-3}-\frac{\left(2x+5\right)\left(x-1\right)}{x^2+2x-3}+\frac{4}{x^2+2x-3}-\frac{x^2+2x-3}{x^2+2x-3}=0\)
\(\Leftrightarrow3x^2+8x-3-\left(2x^2+3x-5\right)+4-\left(x^2+2x-3\right)=0\)
\(\Leftrightarrow3x+9=0\)
\(\Rightarrow x=-3\) (ko thỏa mãn)
Vậy pt vô nghiệm
3) \(\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{4-8x}{x^2-9}\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)-x\left(x+3\right)=4-8x\)
\(\Leftrightarrow x^2-3x-x+3-x^2-3x=4-8x\)
\(\Leftrightarrow x=1\)
Vậy : \(S=\left\{1\right\}\)
a, \(\frac{3}{x-1}=\frac{3x+2}{1-x^2}-\frac{4}{x+1}ĐK:x\ne\pm1\)
\(\Leftrightarrow\frac{3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{3x+2}{\left(1-x\right)\left(x+1\right)}-\frac{4\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow\frac{3x+3}{\left(x-1\right)\left(x+1\right)}=\frac{-3x-2-4x+4}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow3x+3=-7x+2\)
\(\Leftrightarrow10x=-1\Leftrightarrow x=-\frac{1}{10}\)( tmđk )
Vậy tập nghiệm phương trình là : \(S=\left\{-\frac{1}{10}\right\}\)
b, \(x^2-9=\left(x+3\right)\left(1-x\right)\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x+3\right)\left(1-x\right)\)
\(\Leftrightarrow\left(x+3\right)\left(x-3-1+x\right)=0\Leftrightarrow x=-3;x=\frac{1}{2}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-3;\frac{1}{2}\right\}\)
Trả lời:
a, \(\frac{3}{x-1}=\frac{3x+2}{1-x^2}-\frac{4}{x+1}\)\(\left(đkxđ:x\ne\pm1\right)\)
\(\Leftrightarrow\frac{3}{x-1}=\frac{-3x-2}{x^2-1}-\frac{4}{x+1}\)
\(\Leftrightarrow\frac{3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{-3x-2}{\left(x-1\right)\left(x+1\right)}-\frac{4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow3x+3=-3x-2-4x+4\)
\(\Leftrightarrow3x+3=-7x+2\)
\(\Leftrightarrow3x+7x=2-3\)
\(\Leftrightarrow10x=-1\)
\(\Leftrightarrow x=\frac{-1}{10}\)(tm)
Vậy \(S=\left\{\frac{-1}{10}\right\}\)