\(|x-7|\)+2x=\(\dfrac{1}{2}\)
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L
19 tháng 5 2022
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
6 tháng 2 2018
a ) \(\dfrac{1}{x-1}-\dfrac{7}{x+2}=\dfrac{3}{x^2+x-2}\) (1)
ĐKXĐ : x\(\ne1;-2.\)
\(\left(1\right)\Leftrightarrow x+2-7x+7=3\)
\(\Leftrightarrow-6x=-6\)
\(\Leftrightarrow x=1\left(loại\right)\)
Vậy pt vô nghiệm .
b ) \(\dfrac{x^2+2x+1}{x^2+2x+2}+\dfrac{x^2+2x+2}{x^2+2x+3}=\dfrac{7}{6}\)
Đặt \(x^2+2x+1=t\) ta được :
\(\dfrac{t}{t+1}+\dfrac{t+1}{t+2}=\dfrac{7}{6}\)
\(\Leftrightarrow6t^2+12t+6t^2+12t+6=7\left(t^2+3t+2\right)\)
\(\Leftrightarrow5t^2+3t-8=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{8}{5}\end{matrix}\right.\)
Khi t = 1
\(\Leftrightarrow\left(x+1\right)^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Khi \(t=-\dfrac{8}{5}\)
\(\Leftrightarrow\left(x+1\right)^2=-\dfrac{8}{5}\) ( vô lí )
Vậy ............
19 tháng 2 2022
a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)
\(\Leftrightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow39x-84x=1092-73\)
=>-45x=1019
hay x=-1019/45
b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
=>21x+63-14=20x+36-49x+63
=>21x+49=-29x+99
=>50x=50
hay x=1
c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)
=>14x+7-15x-6-21x-63=0
=>-22x-64=0
hay x=-32/11
d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)
=>70x-105-30x-45=84x+63-1785
=>40x-150-84x+1722=0
=>-44x+1572=0
hay x=393/11
TB
2
18 tháng 5 2021
Máy bài này khá dễ chỉ cần suy nghĩ tí là làm được.
Đặt `2x^2-x=a(a>=-1/8)`
`pt<=>1/(a+1)+3/(a+3)=10/(a+7)`
`<=>(a+3)(a+7)+3(a+1)(a+7)=10(a+1)(a+3)`
`<=>a^2+10a+21+3(a^2+8a+7)=10(a^2+4a+3)`
`<=>a^2+3a^2+10a+24a+21+21=10a^2+40a+30`
`<=>4a^2+34a+42=10a^2+40a+30`
`<=>6a^2+6a-12=0`
`<=>a^2+a-2=0`
`a+b+c=0`
`=>a_1=1,a_2=-2(l)`
`a=1=>2x^2-x=1`
`=>2x^2-x-1=0`
`a+b+c=0`
`=>x_1=1,x_1=-1/2`
Vậy `S={1,-1/2}`
18 tháng 5 2021
Đặt \(2x^2-x+1=a\left(a\ge\dfrac{7}{8}\right)\)
PTTT : \(\dfrac{1}{a}+\dfrac{3}{a+2}=\dfrac{10}{a+6}\)
\(\Leftrightarrow\left(a+2\right)\left(a+6\right)+3a\left(a+6\right)=10a\left(a+2\right)\)
\(\Leftrightarrow a^2+2a+6a+12+3a^2+18a=10a^2+20a\)
\(\Leftrightarrow-6a^2+6a+12=0\)
\(\Leftrightarrow\left(a+1\right)\left(a-2\right)=0\)
\(\Leftrightarrow a=2\)
-Thay lại a = 2 ta được : \(2x^2-x-1=0\)
<=> \(\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy ...
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 3 2022
1.
Do \(\lim\limits_{x\rightarrow2}\left(3x-5\right)=1>0\)
\(\lim\limits_{x\rightarrow2}\left(x-2\right)^2=0\)
\(\left(x-2\right)^2>0;\forall x\ne2\)
\(\Rightarrow\lim\limits_{x\rightarrow2}\dfrac{3x-5}{\left(x-2\right)^2}=+\infty\)
2.
\(\lim\limits_{x\rightarrow1^-}\left(2x-7\right)=-5< 0\)
\(\lim\limits_{x\rightarrow1^-}\left(x-1\right)=0\)
\(x-1< 0;\forall x< 1\)
\(\Rightarrow\lim\limits_{x\rightarrow1^-}\dfrac{2x-7}{x-1}=+\infty\)
3.
\(\lim\limits_{x\rightarrow1^+}\left(2x-7\right)=-5< 0\)
\(\lim\limits_{x\rightarrow1^+}\left(x-1\right)=0\)
\(x-1>0;\forall x>1\)
\(\Rightarrow\lim\limits_{x\rightarrow1^+}\dfrac{2x-7}{x-1}=-\infty\)
AN
1
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 3 2023
a.
ĐKXĐ: \(x\ne6\)
\(\dfrac{7}{x-6}=\dfrac{x-6}{7}\)
\(\Leftrightarrow\dfrac{49}{7\left(x-6\right)}=\dfrac{\left(x-6\right)^2}{7\left(x-6\right)}\)
\(\Rightarrow\left(x-6\right)^2=49=7^2\)
\(\Rightarrow\left[{}\begin{matrix}x-6=7\\x-6=-7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=13\\x=-1\end{matrix}\right.\) (thỏa mãn)
b. ĐKXĐ: \(x\ne\dfrac{1}{2}\)
\(\dfrac{2x-1}{8}=\dfrac{-2}{1-2x}\)
\(\Leftrightarrow\dfrac{\left(2x-1\right)^2}{8\left(2x-1\right)}=\dfrac{16}{8\left(2x-1\right)}\)
\(\Rightarrow\left(2x-1\right)^2=16=4^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\) (thỏa mãn)
DN
1
23 tháng 7 2022
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
5 tháng 1 2021
a) ĐKXĐ: x≠-5
Ta có: \(\dfrac{2x-5}{x+5}=4\)
\(\Leftrightarrow2x-5=4\left(x+5\right)\)
\(\Leftrightarrow2x-5=4x+20\)
\(\Leftrightarrow2x-5-4x-20=0\)
\(\Leftrightarrow-2x-25=0\)
\(\Leftrightarrow-2x=25\)
hay \(x=\dfrac{-25}{2}\)(nhận)
Vậy: \(S=\left\{-\dfrac{25}{2}\right\}\)
b) ĐKXĐ: x≠0
Ta có: \(\dfrac{x^2-4}{x}=\dfrac{2x+3}{2}\)
\(\Leftrightarrow2\left(x^2-4\right)=x\left(2x+3\right)\)
\(\Leftrightarrow2x^2-8=2x^2+3x\)
\(\Leftrightarrow2x^2-8-2x^2-3x=0\)
\(\Leftrightarrow-3x-8=0\)
\(\Leftrightarrow-3x=8\)
hay \(x=\dfrac{-8}{3}\)(nhận)
Vậy: \(S=\left\{-\dfrac{8}{3}\right\}\)
c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{2};-5\right\}\)
Ta có: \(\dfrac{2x+3}{2x-1}=\dfrac{x-3}{x+5}\)
\(\Leftrightarrow\left(2x+3\right)\left(x+5\right)=\left(2x-1\right)\left(x-3\right)\)
\(\Leftrightarrow2x^2+10x+3x+15=2x^2-6x-x+3\)
\(\Leftrightarrow2x^2+13x+15=2x^2-7x+3\)
\(\Leftrightarrow2x^2+13x+15-2x^2+7x-3=0\)
\(\Leftrightarrow20x+12=0\)
\(\Leftrightarrow20x=-12\)
hay \(x=-\dfrac{3}{5}\)(nhận)
Vậy: \(S=\left\{-\dfrac{3}{5}\right\}\)
d) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)
Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(x+7\right)\left(6x+1\right)\)
\(\Leftrightarrow6x^2-9x-4x+6=6x^2+x+42x+7\)
\(\Leftrightarrow6x^2-13x+6=6x^2+43x+7\)
\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)
\(\Leftrightarrow-56x-1=0\)
\(\Leftrightarrow-56x=1\)
hay \(x=-\dfrac{1}{56}\)(nhận)
Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)
Lời giải:
$|x-7|=\frac{1}{2}-2x$
$\Rightarrow \frac{1}{2}-2x\geq 0\Rightarrow x\leq \frac{1}{4}$
$\Rightarrow x-7<0\Rightarrow |x-7|=7-x$. Khi đó ta có:
$7-x+2x=\frac{1}{2}$
$7+x=\frac{1}{2}$
$x=\frac{1}{2}-7=\frac{-13}{2}$ (thỏa mãn)
∣x−7∣ + 2x = 1/2
- ( x - 7 ) + 2x = 1/2 ( Tuy nhiên, x<7 )
( x - 7 ) + 2x = 1/2 ( Tuy nhiên, x > hoặc = 7 )
x = -13/2 ( Tuy nhiên, x<7 )
x = 5/2 ( Tuy nhiên, x> hoặc = 7 )
x = -13/2
x thuộc \(\varnothing\)
x = -13/2
Vậy x = -13/2