K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NB
0
14 tháng 2 2018
a) Ta có: \(\frac{x+a}{x+2}+\frac{x-2}{x-a}=2\left(1\right)\)
Với a = 4
Thay vào phương trình (t) ta được:
\(\frac{x+2}{x+2}+\frac{x-2}{x-2}=2\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow x^2-4+x^2-4=2\left(x^2-4\right)\)
\(\Leftrightarrow2x^2=2x^2-8\)
\(\Leftrightarrow0x=-8\)
Vậy phương trình vô nghiệm
b) Nếu x = -1
\(\Rightarrow\frac{-1+a}{-1+2}+\frac{-1-2}{-1-a}=2\)
\(\Leftrightarrow\frac{-1+a}{1}+\frac{-3}{-1-a}=2\)
\(\Leftrightarrow\frac{\left(-1+a\right)\left(-1-a\right)}{-1-a}+\frac{-3}{-1-a}=\frac{2\left(-1-a\right)}{-1-a}\)
\(\Leftrightarrow1+a-a-a^2-3=-2-2a\)
\(\Leftrightarrow-a^2+2a=-2-1+3\)
\(\Leftrightarrow a\left(2-a\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=0\\2-a=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=0\\a=2\end{cases}}}\)
Vậy a = {0;2}
NĂM MỚI VUI VẺ
NV
1
14 tháng 12 2018
DDK : \(x\ge1\)
\(\sqrt{x-1}-\sqrt{5x-1}=\sqrt{3x-2}\)
\(\Leftrightarrow\sqrt{x-1}=\sqrt{3x-2}+\sqrt{5x-1}\)
\(\Rightarrow x-1=3x-2+5x-2+2\sqrt{\left(3x-2\right)\left(5x-1\right)}\)
\(\Leftrightarrow x-1-3x+2-5x+2=2\sqrt{15x^2-3x-10x+2}\)
\(\Leftrightarrow3-7x=2\sqrt{15x^2-13x+2}\)
\(\Rightarrow9-42x+49x^2=4\left(15x^2-13x+2\right)\)
\(\Leftrightarrow9-42x+49x^2=60x^2-52x+8\)
\(\Leftrightarrow11x^2-10x-1=0\)
\(\Leftrightarrow11x^2-11x+x-1=0\)
\(\Leftrightarrow\left(11x+1\right)\left(x-1\right)=0\)
Giải nốt nha .
DN
1
HN
26 tháng 1 2019
\(\frac{x-3}{11}+\frac{x+1}{3}=\frac{x+7}{9}-1\)
\(\Leftrightarrow\frac{9\left(x-3\right)}{99}+\frac{33\left(x+1\right)}{99}=\frac{11\left(x+7\right)}{99}-\frac{99}{99}\)
\(\Leftrightarrow\frac{9\left(x-3\right)+33\left(x+1\right)}{99}=\frac{11\left(x+7\right)-99}{99}\)
\(\Leftrightarrow9\left(x-3\right)+33\left(x+1\right)=11\left(x+7\right)-99\)
\(\Leftrightarrow9x-27+33x+33=11x+77-99\)
\(\Leftrightarrow42x+6=11x-22\Leftrightarrow42x-11x=-6-22\)
\(\Leftrightarrow31x=-28\Leftrightarrow x=-\frac{28}{31}\)
Vậy phương trình có tập nghiệm S={-28/31}
KN
28 tháng 1 2020
\(ĐKXĐ:x\ne-1;x\ne2\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5x+5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow\frac{x-2-5x-5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow x-2-5x-5=15\)
\(\Leftrightarrow-4x=22\Leftrightarrow x=\frac{-11}{2}\)
Vậy \(S=\left\{\frac{-11}{2}\right\}\)
HN
28 tháng 1 2020
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(x-2\right)}\left(ĐKXĐ:x\ne-1;x\ne2\right)\)
\(\Leftrightarrow\frac{1\left(x-2\right)-5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\frac{x-2-5x-5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\frac{-4x-7}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow-4x-7=15\)
\(\Leftrightarrow-4x=22\)
\(\Leftrightarrow x=22:\left(-4\right)\)
\(\Leftrightarrow x=\frac{-22}{4}=\frac{-11}{2}\)
Vậy tập nghiệm \(S=\left\{\frac{-11}{2}\right\}\)
TB
1
27 tháng 8 2017
\(\sqrt{x+5}=1+\sqrt{x}\)
ĐKXĐ : \(x\ge0\)
\(pt\Leftrightarrow x+5=\left(1+\sqrt{x}\right)^2\)
\(\Leftrightarrow x+5=x+2\sqrt{x}+1\)
\(\Leftrightarrow x+5-x-2\sqrt{x}-1=0\)
\(\Leftrightarrow-2\sqrt{x}+4=0\)
\(\Leftrightarrow\sqrt{x}=2\Rightarrow x=4\)(TMĐKXĐ)
TD
0
KS
2
21 tháng 7 2019
\(\frac{\left(x-2\right)^2}{3}-\frac{2x-1}{4}=4-\frac{\left(2x-3\right)^2}{6}.\)
\(\Rightarrow\frac{4\left(x-2\right)^2}{12}-\frac{3\left(2x-1\right)^2}{12}=\frac{48}{12}-\frac{2\left(2x-3\right)^2}{12}\)
\(\Rightarrow4\left(x^2-4x+4\right)-3\left(4x^2-4x+1\right)=48-2\left(4x^2-12x+9\right)\)
\(\Rightarrow4x^2-16x+16-12x^2+12x-3=48-8x^2+24x-18\)
\(\Rightarrow-16x+12x+16-3=24x+48-18\)
\(\Rightarrow28x=-17\Leftrightarrow x=-\frac{17}{28}\)