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a) \(\dfrac{2x+3}{24}=\dfrac{3x-1}{32}\\ =>32\left(2x+3\right)=24\left(3x-1\right)\\ =>64x+96=72x-24\\ =>72x-64x=24+96\\ =>8x=120\\ =>x=120:8\\ =>x=15\)
b) \(\dfrac{13x-2}{2x+5}=\dfrac{76}{17}\\=>76\left(2x+5\right)=17\left(13x-2\right)\\ =>152x+380=221x-34\\ =>221x-152x=34+380\\ =>69x=414\\ =>x=414:69\\ =>x=6\)
a.
\(\dfrac{2x+3}{24}=\dfrac{3x-1}{32}\)
\(\Leftrightarrow\dfrac{4\left(2x+3\right)}{4.24}=\dfrac{3\left(3x-1\right)}{32.3}\)
\(\Leftrightarrow\dfrac{8x+12}{96}=\dfrac{9x-3}{96}\)
\(\Leftrightarrow8x+12=9x-3\)
\(\Leftrightarrow9x-8x=12+3\)
\(\Leftrightarrow x=15\)
b.
ĐKXĐ: \(x\ne-\dfrac{5}{2}\)
\(\dfrac{13x-2}{2x+5}=\dfrac{76}{17}\)
\(\Leftrightarrow\dfrac{17\left(13x-2\right)}{17\left(2x+5\right)}=\dfrac{76\left(2x+5\right)}{17\left(2x+5\right)}\)
\(\Rightarrow17\left(13x-2\right)=76\left(2x+5\right)\)
\(\Leftrightarrow221x-34=152x+380\)
\(\Leftrightarrow69x=414\)
\(\Leftrightarrow x=6\)
1. Thay x = 1 vào đa thức f (x) = ax2 + bx + c . Ta có :
f ( x ) = a.12 + b.1 + c
= a + b + c
= 0
Vậy x = 1 là nghiệm của f ( x )
Bài 1 :
Giả sử x = 1 là nghiệm của đa thức f (x) = ax2 + bx + c
=> f (x) = a . 12 + b . 1 + c = 0
<=> f(x) = a + b + c = 0
Vậy nếu a + b + c = 0 thì x = 1 là nghiệm của đa thứ f (x)
Bài 2 :
a) \(\left(x-2\right)\left(2x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=4\end{cases}}\)
Vậy nghiệm của đa thức là x=2 hoặc x=4
b) \(\left(3x-9\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-9=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)
Vậy .................
c) \(\left(x-3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x-3=0\left(x^2+1>0\right)\)
\(\Leftrightarrow x=3\)
Vậy .............
d) \(\left(x^2+2\right)\left(x^2-3\right)=0\)
\(\Leftrightarrow x^2-3=0\left(x^2+2>0\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Vậy...............
Bài 1:
a: \(\left|x-\dfrac{1}{2}\right|+\dfrac{1}{2}=x\)
=>\(\left|x-\dfrac{1}{2}\right|=x-\dfrac{1}{2}\)
=>\(x-\dfrac{1}{2}>=0\)
=>\(x>=\dfrac{1}{2}\)
b: \(\left|1-3x\right|+1=3x\)
=>\(\left|1-3x\right|=3x-1\)
=>\(1-3x< =0\)
=>3x-1>=0
=>3x>=1
=>\(x>=\dfrac{1}{3}\)
Bài 2:
a: \(C=\left|5-x\right|+x=\left|x-5\right|+x\)
TH1: x>=5
\(C=x-5+x=2x-5\)
TH2: x<5
C=5-x+x=5
b: D=|2x-1|-x
TH1: x>=1/2
\(D=2x-1-x=x-1\)
TH2: \(x< \dfrac{1}{2}\)
D=1-2x-x=1-3x
a: \(3x-\left|2x+1\right|=2\)
\(\Leftrightarrow\left|2x+1\right|=3x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x-2\right)^2-\left(2x+1\right)^2=0\\x>=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(3x-2-2x-1\right)\left(3x-2+2x+1\right)=0\\x>=\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)\left(5x-1\right)=0\\x>=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow x=3\)
e: Ta có: \(2n-3⋮n+1\)
\(\Leftrightarrow2n+2-5⋮n+1\)
\(\Leftrightarrow n+1\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{0;-2;4;-6\right\}\)
\(a,\frac{34}{5}:\frac{8}{5}=0,25:x\)
\(\frac{17}{4}=\frac{1}{4}:x\)
\(x=\frac{1}{4}:\frac{17}{4}\)
\(x=\frac{1}{17}\)
\(b,2x+\frac{3}{24}=3x-\frac{1}{32}\)
\(2x-3x=\frac{-1}{32}-\frac{1}{8}\)
\(-x=\frac{-5}{32}\)
\(x=\frac{5}{32}\)
\(13x-\frac{2}{2}x+5=\frac{76}{17}\)
\(12x=\frac{-9}{17}\)
\(x=\frac{-3}{68}\)
a,\(\frac{34}{5}\div\frac{8}{5}=0,25\div x\)
\(\frac{17}{4}=0,25\div x\)
\(x=17\)
b,\(2x+\frac{3}{24}=3x-\frac{1}{32}\)
\(\Leftrightarrow2x-3x=-\frac{1}{32}-\frac{3}{24}\)
\(\Leftrightarrow2x-3x=-\frac{5}{32}\)
\(\Leftrightarrow-x=-\frac{5}{32}\)
\(\Leftrightarrow x=\frac{5}{32}\)