Giúp mk câu này với nhanh dùm : 4x-(x+\(\dfrac{1}{2}\))=2x-(\(\dfrac{1}{2}\)-5) trình bày ra nhé , cho mk thank you trước . Hihhihii...
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17 tháng 1 2023
\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\left(ĐKXĐ:x\ne5\right)\)
\(\Rightarrow3\left(4x-3\right)=29\left(x-5\right)\)
\(\Leftrightarrow12x-9=29x-145\)
\(\Leftrightarrow12x-9-29x+145=0\)
\(\Leftrightarrow-17x+136=0\)
\(\Leftrightarrow-17x=-136\)
\(\Leftrightarrow x=8\left(tm\right)\)
Vậy \(S=\left\{8\right\}\)
\(2,\dfrac{2x-1}{5-3x}=2\left(ĐKXĐ:x\ne\dfrac{5}{3}\right)\)
\(\Rightarrow2x-1=2\left(5-3x\right)\)
\(\Leftrightarrow2x-1=10-6x\)
\(\Leftrightarrow2x-1-10+6x=0\)
\(\Leftrightarrow8x-11=0\)
\(\Leftrightarrow8x=11\)
\(\Leftrightarrow x=\dfrac{11}{8}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{11}{8}\right\}\)
\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\left(ĐKXĐ:x\ne1\right)\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2\left(x-1\right)}{x-1}+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2x-2}{x-1}+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{3x-2}{x-1}\)
\(\Rightarrow4x-5=3x-2\)
\(\Leftrightarrow4x-5-3x+2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\left(tm\right)\)
Vậy \(S=\left\{3\right\}\)
\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne-5\right)\)
\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2+15x+25}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow\dfrac{15x+25}{2x\left(x+5\right)}=0\)
\(\Rightarrow15x+25=0\)
\(\Leftrightarrow15x=-25\)
\(\Leftrightarrow x=\dfrac{-5}{3}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{-5}{3}\right\}\)
Y
17 tháng 1 2023
\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-29\left(x-5\right)}{3\left(x-5\right)}=0\)
\(\Leftrightarrow12x-9-29x+145=0\)
\(\Leftrightarrow-17x=-136\)
\(\Leftrightarrow x=8\)
\(2,\dfrac{2x-1}{5-3x}=2\)
\(\Leftrightarrow\dfrac{2x-1-2\left(5-3x\right)}{5-3x}=0\)
\(\Leftrightarrow2x-1-10+6x=0\)
\(\Leftrightarrow8x=11\)
\(\Leftrightarrow x=\dfrac{11}{8}\)
\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5-2\left(x-1-x\right)}{x-1}=0\)
\(\Leftrightarrow4x-5-2x+2+2x=0\)
\(\Leftrightarrow4x=3\)
\(\Leftrightarrow x=\dfrac{3}{4}\)
\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)
\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)-2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow2x^2+10x+5x+25-2x^2=0\)
\(\Leftrightarrow15x=-25\)
\(\Leftrightarrow x=-\dfrac{5}{3}\)
6 tháng 4 2022
Câu 1 :
a. \(4x-5=23\\ \Leftrightarrow4x=23+5\\ \Leftrightarrow4x=28\\ \Leftrightarrow x=7\)
b.
|-2x|=5x+14
Nếu - 2x > 0 => x < 0 thì |-2x|= - 2x, ta có pt: -2x = 5x+14
<=> - 2x = 5x + 14
<=> - 2x - 5x = 14
<=> - 7x = 14
<=> x = - 2 (thoã mãn)
Nếu - 2x < 0 => x > 0 thì |-2x|= = -(- 2x) = 2x.
Ta có pt: 2x = 5x + 14
<=> - 3x = 14
<=> x = \(-\dfrac{14}{3}\)
Vậy pt có nghiệm x = - 2
c) \(\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{x^2-1}\\ ĐKXĐ:x\ne1;x\ne-1\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{1\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+2}{\left(x-1\right)\left(x+1\right)}\\ \Leftrightarrow x^2+x+x+1-x+1=x^2+2\\ \Leftrightarrow x^2+x+x-x-x^2=2-1-1\\ \Leftrightarrow x=0\left(nhận\right)\)
Y
6 tháng 4 2022
\(a,4x-5=23\)
\(\Leftrightarrow4x=23+5\)
\(\Leftrightarrow4x=28\)
\(\Leftrightarrow x=7\)
\(b,\left|-2x\right|=5x+14\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5x+14\\2x=-5x-14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x-14=0\\7x+14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=14\\7x=-14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{14}{3}\\x=-2\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{14}{3};-2\right\}\)
\(c,\Leftrightarrow\dfrac{\left(x+1\right)\left(x+1\right)-x+1-x^2-2}{x^2-1}=0\)
\(\Leftrightarrow x^2+x+x+1-x+1-x^2-2=0\)
\(\Leftrightarrow x=0\)
Vậy \(S=\left\{0\right\}\)
1 tháng 2 2017
x-2x+5y-12=0
y(x+5)-2(x+5)-2=0
(y-2)(x+5)=2
xong thu TH nha
k cho toi di Hien beo ne
17 tháng 1 2022
\(\dfrac{x}{2}-\dfrac{2}{9}=\dfrac{3}{2}-\left(\dfrac{-5}{4}\right)\\ \Rightarrow\dfrac{x}{2}-\dfrac{2}{9}=\dfrac{11}{4}\\ \Rightarrow\dfrac{x}{2}=\dfrac{11}{4}+\dfrac{2}{9}\\ \Rightarrow\dfrac{x}{2}=\dfrac{107}{36}\)
\(\Rightarrow x=\dfrac{107}{36}\times2\\ \Rightarrow x=\dfrac{107}{18}\)
AH
Akai Haruma
Giáo viên
13 tháng 12 2021
Lời giải:
a.
\(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)
\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)
\(\frac{1}{6-3x}=\frac{x+2}{3(2-x)}=\frac{2(x+2)^2}{6(2-x)(2+x)}=\frac{-2(x+2)^2}{6(x^2-4)}\)
b.
\(\frac{1}{x+2}=\frac{x(2-x)}{x(x+2)(2-x)}=\frac{x(2-x)}{x(4-x^2)}\)
\(\frac{8}{2x-x^2}=\frac{8(x+2)}{(x+2)x(2-x)}=\frac{8(x+2)}{x(4-x^2)}\)
c.
\(\frac{4x^2-3x+5}{x^3-1}\)
\(\frac{1-2x}{x^2+x+1}=\frac{(1-2x)(x-1)}{(x-1)(x^2+x+1)}=\frac{-2x^2+3x-1}{x^3-1}\)
\(-2=\frac{-2(x^3-1)}{x^3-1}\)
4 tháng 7 2021
a)Pt \(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\dfrac{1}{3}+\dfrac{1}{2}\)
\(\Leftrightarrow\left|2x-1\right|=\dfrac{5}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\dfrac{5}{6}\\2x-1=-\dfrac{5}{6}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{12}\\x=\dfrac{1}{12}\end{matrix}\right.\)
Vậy...
b)Đk:\(x\ge3\)
Pt \(\Leftrightarrow\sqrt{x-3}\left(x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\x-4=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=4\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)
Vậy...
c)Đk:\(x\ge1\)
\(x+\sqrt{x-1}=13\)
\(\Leftrightarrow\sqrt{x-1}=13-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}13-x\ge0\\x-1=x^2-26x+169\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\x^2-27x+170=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\x^2-17x-10x+170=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\\left(x-17\right)\left(x-10\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\\left[{}\begin{matrix}x=17\\x=10\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow x=10\) (tm)
Vậy...
30 tháng 6 2021
a) Ta có: \(A=\left(\dfrac{x-\sqrt{x}+2}{x-\sqrt{x}-2}-\dfrac{x}{x-2\sqrt{x}}\right):\dfrac{1-\sqrt{x}}{2-\sqrt{x}}\)
\(=\left(\dfrac{x-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)
\(=\dfrac{x-\sqrt{x}+2-x-\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)
\(=\dfrac{-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=-\dfrac{2}{\sqrt{x}+1}\)
b) Ta có: \(B=\left(\dfrac{1}{2\sqrt{x}+1}+\dfrac{1}{2\sqrt{x}-1}\right):\dfrac{1}{1-4x}\)
\(=\dfrac{2\sqrt{x}-1+2\sqrt{x}+1}{\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\cdot\dfrac{-\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{1}\)
\(=-4\sqrt{x}\)
15 tháng 2 2022
\(Đk:\) \(x\ne1,x\ne2,x\ne3\)
\(\Rightarrow\dfrac{x+4}{\left(x-2\right)\left(x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(x-1\right)}=\dfrac{2x+5}{\left(x-3\right)\left(x-1\right)}\)
\(\Rightarrow\dfrac{\left(x+4\right)\cdot\left(x-3\right)+\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)\left(x-3\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(x-3\right)\left(x-1\right)\left(x-2\right)}\)
\(\Rightarrow x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)
\(\Rightarrow0x-14=x-10\)
\(\Rightarrow x=-4\left(tmđk\right)\)
Phá ngoặc ra là ok rồi mà:
\(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)
\(\Rightarrow4x-x-\dfrac{1}{2}=2x-\dfrac{1}{2}+5\)
\(\Rightarrow3x-\dfrac{1}{2}=2x+\dfrac{9}{2}\)
\(\Rightarrow3x-\dfrac{1}{2}-\dfrac{9}{2}=2x\)
\(\Rightarrow3x-5=2x\)
\(\Rightarrow x=5\)
Ta :có
\(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)
\(\Leftrightarrow4x-2x=\left(\dfrac{1}{2}-5\right)+\left(x+\dfrac{1}{2}\right)\)
\(\Leftrightarrow2x=-4+x\)
\(\Leftrightarrow2x-x=-4\)
\(\Rightarrow x=-4\)