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a) Đúng
b)Đúng
c)Sai vì nghiệm không thỏa mãn ĐKXĐ
d)Sai vì có 1 nghiệm không thỏa mãn ĐKXĐ
Câu 1: D. \(\frac{1}{2}-4x=0\)
Câu 2: C. 2x - 1 = x
Câu 3: D. S = {-9}
# Chúc bạn học tốt #
a. \(5x-10=0\)
\(\Leftrightarrow5x=10\)
\(\Leftrightarrow x=2\)
Vậy \(S=\left\{2\right\}\)
b. \(\dfrac{2}{x+1}-\dfrac{3}{x-2}=\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
ĐKXĐ: \(x\ne-1;x\ne2\)
\(\Leftrightarrow\dfrac{2\left(x-2\right)}{x+1\left(x-2\right)}-\dfrac{3\left(x+1\right)}{x-2\left(x+1\right)}=\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow2\left(x-2\right)-3\left(x+1\right)=2x-6\)
\(\Leftrightarrow2x-4-3x-3=2x-6\)
\(\Leftrightarrow2x-4-3x-3-2x+6=0\)
\(\Leftrightarrow-3x-1=0\)
\(\Leftrightarrow-3x=1\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Vậy \(S=\left\{-\dfrac{1}{3}\right\}\)
c) \(3x-5\ge-7\) (3)
\(\Leftrightarrow3x\ge-7+5\)
\(\Leftrightarrow3x\ge-2\)
\(\Leftrightarrow x\ge-\dfrac{2}{3}\)
Vậy tập nghiệm của BPT (3) là \(x\ge-\dfrac{2}{3}\)
d) \(3x-1=0\)
\(\Leftrightarrow3x=1\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
Vậy \(S=\left\{\dfrac{1}{3}\right\}\)
a. 5x - 10 = 0
⇔ 5x = 10
⇔ x = 2
Vậy S ={2}.
b.\(\dfrac{2}{x+1}\) - \(\dfrac{3}{x-2}\) = \(\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
⇔\(\dfrac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}\) - \(\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}\) = \(\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
⇔ 2(x - 2) - 3(x+1) = 2x - 6
⇔ 2x - 4 - 3x -3 = 2x - 6
⇔ 2x - 2x - 3x = -6 + 4 + 3
⇔ -3x = 1
⇔ x = \(-\dfrac{1}{3}\)
b: =>(2x-1)(2x-1+4-2x)=0
=>3(2x-1)=0
=>2x-1=0
=>x=1/2
c: =>(x+1)(x^2-x+1)-x(x+1)=0
=>(x+1)(x-1)^2=0
=>x=1 hoặc x=-1
e: =>(2x-1)(2x+1)=0
=>x=1/2 hoặc x=-1/2
h: =>x[(x^2-5)^2-4]=0
=>x(x^2-7)(x^2-3)=0
=>\(x\in\left\{0;\pm\sqrt{7};\pm\sqrt{3}\right\}\)
k: =>(x-1)(5x+3-3x+8)=0
=>(x-1)(2x+11)=0
=>x=1 hoặc x=-11/2
l: =>x^2(x+1)+(x+1)=0
=>(x+1)(x^2+1)=0
=>x+1=0
=>x=-1
ĐKXĐ:\(x\ne\pm2;x\ne-3;x\ne0\)
\(P=1+\frac{x-3}{x^2+5x+6}\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left(\frac{2}{x-2}-\frac{x}{x^2-4}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\cdot\frac{2x+4-x-x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\frac{8\left(x-3\right)}{\left(x+2\right)^2\left(x+3\right)\left(x-2\right)}\)
Đề sai à ??
d)\(x^2-y^2+2x-4y-10=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)-\left(y^2+4y+4\right)=7\)
\(\Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\)
\(\Leftrightarrow\left(x-y-1\right)\left(x+y+3\right)=7\)
Mà x,y nguyên dương\(\Rightarrow x-y-1< x+y+3\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-y-1=1\\x+y+3=7\end{matrix}\right.\\\left\{{}\begin{matrix}x-y-1=-7\\x+y+3=-1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
Mạn phép ko chép lại đề :
b) \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}-x^2-2-\dfrac{1}{x^2}\right)=\left(x+4\right)^2\)
⇔ \(8\left(x+\dfrac{1}{x}\right)^2-8\left(x^2+\dfrac{1}{x^2}\right)=\left(x+4\right)^2\) ( x # 0)
⇔ \(8\left(x^2+2+\dfrac{1}{x^2}-x^2-\dfrac{1}{x^2}\right)=\left(x+4\right)^2\)
⇔ ( x + 4)2 = 16
⇔ x2 + 8x + 16 = 16
⇔ x( x + 8) = 0
⇔ x = 0 ( KTM) hoặc : x = - 8 ( TM)
KL.....
Giải:
a) \(3x+5=14\)
\(\Leftrightarrow3x=14-5=9\)
\(\Leftrightarrow x=\dfrac{9}{3}=3\)
Vậy ...
b) \(\left(x+3\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy ...
c) \(\dfrac{x-1}{x-2}+\dfrac{x+3}{x-4}=\dfrac{2}{\left(x-2\right)\left(4-x\right)}\) (1)
ĐKXĐ: \(x\ne2;x\ne4\)
\(\left(1\right)\Leftrightarrow\dfrac{x-1}{x-2}+\dfrac{-x-3}{4-x}=\dfrac{2}{\left(x-2\right)\left(4-x\right)}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(4-x\right)}{\left(x-2\right)\left(4-x\right)}+\dfrac{\left(-x-3\right)\left(x-2\right)}{\left(4-x\right)\left(x-2\right)}=\dfrac{2}{\left(x-2\right)\left(4-x\right)}\)
\(\Rightarrow\left(x-1\right)\left(4-x\right)+\left(-x-3\right)\left(x-2\right)=2\)
\(\Leftrightarrow\left(x-1\right)\left(4-x\right)-\left(x+3\right)\left(x-2\right)=2\)
\(\Leftrightarrow4x-4-x^2+x-\left(x^2+3x-2x-6\right)=2\)
\(\Leftrightarrow4x-4-x^2+x-x^2+3x-2x-6=2\)
Tự triển khai và tìm ra nghiệm của phương trình.
d) \(\left|2x-1\right|=x+4\) (2)
TH1: \(2x-1\ge0\Leftrightarrow x\ge\dfrac{1}{2}\)
\(\left(2\right)\Leftrightarrow2x-1=x+4\)
\(\Leftrightarrow2x-x=4+1\)
\(\Leftrightarrow x=5\) (thỏa mãn)
TH2: \(2x-1< 0\Leftrightarrow x< \dfrac{1}{2}\)
\(\left(2\right)\Leftrightarrow-2x+1=x+4\)
\(\Leftrightarrow-2x-x=4-1\)
\(\Leftrightarrow-3x=3\)
\(\Leftrightarrow x=-1\) (thỏa mãn)
Vậy ...
$a) 3x + 5 = 14$
$\Leftrightarrow 3x = 14 - 5$
$\Leftrightarrow 3x = 9$
$\Leftrightarrow x = \frac{9}{3}$
$\Leftrightarrow x = 3$
Vậy tập nghiệm của pt: S = {3}
$b) (x + 3)(2x - 5) = 0$
$\Leftrightarrow x + 3 = 0 hoặc 2x - 5 = 0$
$\Leftrightarrow x = - 3 hoặc 2x = 5$
$\Leftrightarrow x = - 3 hoặc x = \frac{5}{2}$
Vậy tập nghiệm của pt: S = {$- 3$; $\frac{5}{2}$}
C
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