Tìm x biết :
\(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\)
Tìm x, biết:
a) \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\)
b) \(\dfrac{x+1}{2x+1}=\dfrac{0,5x+2}{x+3}\)
a/ đk: x khác -7/5 ; x khác -1/5
pt <=> \(\dfrac{\left(3x+2\right)\left(5x+1\right)}{\left(5x+7\right)\left(5x+1\right)}=\dfrac{\left(3x-1\right)\left(5x+7\right)}{\left(5x+7\right)\left(5x+1\right)}\)
\(\Rightarrow15x^2+13x+2=15x^2+16x-7\)
\(\Leftrightarrow15x^2+13x-15x^2-16x^2=-7-2\)
\(\Leftrightarrow-3x=-9\Leftrightarrow x=3\left(tm\right)\)
vậy x = 3
b/ đk: x khác -1/2; x khác -3
pt <=> \(\dfrac{\left(x+1\right)\left(x+3\right)}{\left(2x+1\right)\left(x+3\right)}=\dfrac{\left(0,5x+2\right)\left(2x+1\right)}{\left(2x+1\right)\left(x+3\right)}\)
\(\Rightarrow x^2+4x+3=x^2+4,5x+2\)
\(\Leftrightarrow x^2+4x-x^2-4,5x=2-3\)
\(\Leftrightarrow-0,5x=-1\Leftrightarrow x=2\left(tm\right)\)
vậy x = 2
a) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}=\dfrac{\left(3x+2\right)-\left(3x-1\right)}{\left(5x+7\right)-\left(5x+1\right)}=\dfrac{3}{6}=\dfrac{1}{2}\)
\(\Rightarrow2\left(3x+2\right)=5x+7\)
\(\Rightarrow6x+4=5x+7\)
\(\Leftrightarrow x=3\)
Vậy x = 3
b) Ta có: \(\dfrac{0,5x+2}{x+3}=\dfrac{2\left(0,5x+2\right)}{2\left(x+3\right)}=\dfrac{x+4}{2x+6}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x+1}{2x+1}=\dfrac{0,5x+2}{x+3}=\dfrac{x+4}{2x+6}=\dfrac{\left(x+4\right)-\left(x+1\right)}{\left(2x+6\right)-\left(2x+1\right)}=\dfrac{3}{5}\)
\(\Rightarrow5\left(x+1\right)=3\left(2x+1\right)\)
\(\Rightarrow5x+5=6x+3\)
\(\Leftrightarrow x=2\)
Giải PT sau:
a, 3x - 7 = 0
b, 8 - 5x = 0
c, 3x - 2 = 5x + 8
d, \(\dfrac{3x-2}{3}\) = \(\dfrac{1-x}{2}\)
e, ( 5x + 1)(x - 3) = 0
f, (x + 1)(2x - 3) = 0
g, 4x(x + 3) - 5(x + 3) = 0
h, 8(x - 6) - 2x(6 - x) = 0
i, \(\dfrac{2}{x-1}\) + \(\dfrac{1}{x}\) = \(\dfrac{2x+5}{x^2-x}\)
k, \(\dfrac{3}{x+2}\) - \(\dfrac{2}{x-2}\) = \(\dfrac{2-x}{x^2-4}\)
m, \(\dfrac{3}{x}\) - \(\dfrac{2}{x-3}\) = \(\dfrac{4-x}{x^2-3}\)
n,\(\dfrac{3}{2x+10}\)+ \(\dfrac{2x}{x^2-25}\) = \(\dfrac{3}{x-5}\)
u, \(\dfrac{2}{x+3}\) - \(\dfrac{3}{x-2}\) = \(\dfrac{x+4}{\left(x+3\right)\left(x-2\right)}\)
a, 3x - 7 = 0
<=> 3x = 7
<=> x = 7/3
b, 8 - 5x = 0
<=> -5x = -8
<=> x = 8/5
c, 3x - 2 = 5x + 8
<=> -2x = 10
<=> x = -5
e) Ta có: \(\left(5x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{5};3\right\}\)
`a ) 3x - 7 = 0`
`\(\Leftrightarrow \) 3x = 7`
`\(\Leftrightarrow \) x = 7/3`
Vậy `S = {-7/3}`
tìm x biết :
a. \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\)
b. \(\dfrac{x+1}{2x+1}=\dfrac{0,5x+2}{x+3}\)
\(\dfrac{3x+2}{5x-7}=\dfrac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x-7\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2-21x-5x-7\)
\(\Leftrightarrow15x^2+13x+2=15x^2-26x-7\)
\(\Leftrightarrow15x^2-15x^2-13x-2=26x-7\)
\(\Leftrightarrow-13x-2=26x-7\)
\(\Leftrightarrow26x+13x=7+2\)
\(\Leftrightarrow39x=9\Leftrightarrow x=\dfrac{3}{13}\)
b tương tự
a) \(\dfrac{5x-2}{3}+x=1+\dfrac{5-3x}{2}\)
b) \(\dfrac{\left(3x-1\right)\left(x+2\right)}{3}-\dfrac{2x^2+1}{2}=\dfrac{11}{2}\)
c) \(x-\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)
d) \(\dfrac{x-1}{2}+\dfrac{x-1}{3}-\dfrac{x-1}{6}=2\)
a, \(\Rightarrow10x-4+6x=6+15-9x\Leftrightarrow7x=25\Leftrightarrow x=\dfrac{25}{7}\)
b, \(\Rightarrow2\left(3x^2+5x-2\right)-6x^2-3=33\Leftrightarrow10x-7=33\Leftrightarrow x=4\)
c, \(\Rightarrow12x-10x-4=21-9x\Leftrightarrow11x=25\Leftrightarrow x=\dfrac{25}{11}\)
d, \(\Rightarrow3x-3+2x-2-x+1=12\Leftrightarrow4x=16\Leftrightarrow x=4\)
\(\dfrac{5x-2}{3}+x=1+\dfrac{5-3x}{2}\)
\(\Leftrightarrow\dfrac{5x-2+3x}{3}=\dfrac{2+5-3x}{2}\)
\(\Leftrightarrow\dfrac{8x-2}{3}=\dfrac{7-3x}{2}\)
\(\Leftrightarrow16x-4=21-9x\)
\(\Leftrightarrow16x+9x=21+4\)
\(\Leftrightarrow25x=25\)
\(\Leftrightarrow x=1\)
\(a,\dfrac{5x-2}{3}+x=1+\dfrac{-3x+5}{2}\)
\(2\left(5x-2\right)+6x=-9x+21\)
\(16x+9x=21+4\)
\(25x=25\)
\(x=1\)
\(b,\dfrac{3x^2+5x-2}{3}-\dfrac{2x^2+1}{2}=\dfrac{11}{2}\)
\(\dfrac{6x^2=10x-4-6x^2-3}{6}=\dfrac{11}{2}\)
\(\dfrac{10x-4-3}{6}=\dfrac{11}{2}\)
\(\dfrac{10x-7}{6}=\dfrac{11}{2}\)
\(10x=33+7\)
\(x=4\)
Tìm x biết :
1) \(\dfrac{-7}{x+1}=\dfrac{-x-1}{\dfrac{4}{7}}\)
2) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
3) \(\dfrac{x+1}{2x+1}=\dfrac{0,5x+2}{x+3}\)
4) \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\)
1: \(\Leftrightarrow\left(x+1\right)^2=4\)
=>x+1=2 hoặc x+1=-2
=>x=1 hoặc x=-3
2: \(\Leftrightarrow7x-21=5x+25\)
=>2x=46
=>x=23
3: \(\Leftrightarrow x^2+4x+3=x^2+0.5x+4x+2\)
=>4,5x+2=4x+3
=>x=1
giải các phương trình sau
1, \(\dfrac{3}{x-3}+\dfrac{4}{x+3}=\dfrac{3x-7}{x^2-9}\)
2, \(\dfrac{3}{x-4}-\dfrac{4}{x+4}=\dfrac{3x-4}{x^2-16}\)
3, \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)
1: Ta có: \(\dfrac{3}{x-3}+\dfrac{4}{x+3}=\dfrac{3x-7}{x^2-9}\)
\(\Leftrightarrow\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}+\dfrac{4x-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-7}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(3x+9+4x-12=3x-7\)
\(\Leftrightarrow4x=-7+12-9=-4\)
hay \(x=-1\left(nhận\right)\)
2: Ta có: \(\dfrac{3}{x-4}-\dfrac{4}{x+4}=\dfrac{3x-4}{x^2-16}\)
\(\Leftrightarrow\dfrac{3x+12}{\left(x-4\right)\left(x+4\right)}-\dfrac{4x-16}{\left(x+4\right)\left(x-4\right)}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\)
Suy ra: \(3x+12-4x+16=3x-4\)
\(\Leftrightarrow28-4x=-4\)
\(\Leftrightarrow4x=32\)
hay \(x=8\left(tm\right)\)
3: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)
Suy ra: \(5x^2-12+3x+3=5x^2-5x\)
\(\Leftrightarrow3x-9+5x=0\)
\(\Leftrightarrow8x=9\)
hay \(x=\dfrac{9}{8}\left(nhận\right)\)
1).(4-3x)(10-5x)=0 2).(7-2x)(4+8x)=0 3).(9-7x)(11-3x)=0
4).(7-14x)(x-2)=0 5).(\(\dfrac{7}{8}\)-2x)(3x+\(\dfrac{1}{3}\))=0 6).3x-2x\(^2\)
7).5x+10x\(^2\)
1.
<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)
2.
<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
3.
<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)
4.
<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)
5.
<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)
6,7. ko đủ điều kiện tìm
1. Giải các BPT
a) \(\dfrac{5x^2-3x}{5}+\dfrac{3x+1}{4}< \dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\)
b)\(\dfrac{5x-20}{3}-\dfrac{2x^2+x}{2}\ge\dfrac{x\left(1-3x\right)}{3}-\dfrac{5x}{4}\)
c) (x+3)2\(\le\)x2-7
\(\text{a) }\dfrac{5x^2-3x}{5}+\dfrac{3x+1}{4}< \dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\\ \Leftrightarrow4\left(5x^2-3x\right)+5\left(3x+1\right)< 10x\left(2x+1\right)-15\\ \Leftrightarrow20x^2-12x+15x+5< 20x^2+10x-15\\ \Leftrightarrow20x^2+3x-20x^2-10x< -15-5\\ \Leftrightarrow-7x< -20\\ \Leftrightarrow x>\dfrac{20}{7}\)
Vậy bất phương trình có nghiệm \(x>\dfrac{20}{7}\)
\(\text{b) }\dfrac{5x-20}{3}-\dfrac{2x^2+x}{2}\ge\dfrac{x\left(1-3x\right)}{3}-\dfrac{5x}{4}\\ \Leftrightarrow4\left(5x-20\right)-6\left(2x^2+x\right)\ge4x\left(1-3x\right)-15x\\ \Leftrightarrow20x-80-12x^2-6x\ge4x-12x^2-15x\\ \Leftrightarrow-12x^2+14x+12x^2+11x\ge80\\ \Leftrightarrow25x\ge80\\ \Leftrightarrow x\ge\dfrac{16}{5}\)
Vậy bất phương trình có nghiệm \(x\ge\dfrac{16}{5}\)
\(\text{c) }\left(x+3\right)^2\le x^2-7\\ \Leftrightarrow x^2+6x+9\le x^2-7\\ \Leftrightarrow x^2+6x-x^2\le-7-9\\ \Leftrightarrow6x\le-16\\ \Leftrightarrow x\le-\dfrac{8}{3}\)
Vậy bất phương trình có nghiệm \(x\le-\dfrac{8}{3}\)
Quy đồng mẫu thức:
a) \(\dfrac{3x+5}{x^2-5x}+\dfrac{25-x}{25-5x}\)
b) \(\dfrac{x+1}{x+3}+\dfrac{x-7}{x^2+x-6}+\dfrac{1}{x-2}\)
\(a,=\dfrac{15x+25-25x+x^2}{5x\left(x-5\right)}=\dfrac{\left(x-5\right)^2}{5x\left(x-5\right)}=\dfrac{x-5}{5x}\\ b,=\dfrac{x^2-x-2+x-7+x+3}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2+x-6}{x^2+x-6}=1\)
\(a,\dfrac{3x+5}{x^2-5x}+\dfrac{25-x}{25-5x}\)
\(=\dfrac{3x+5}{x\left(x-5\right)}+\dfrac{25-x}{5\left(5-x\right)}\)
\(=\dfrac{-3x-5}{x\left(5-x\right)}+\dfrac{25-x}{5\left(5-x\right)}\)
\(=\dfrac{5\left(-3x-5\right)}{5x\left(5-x\right)}+\dfrac{x\left(25-x\right)}{5x\left(5-x\right)}\)
\(=\dfrac{-15x-25+25x-x^2}{5x\left(5-x\right)}\)
\(=\dfrac{10x-25-x^2}{5x\left(5-x\right)}\)
\(=\dfrac{-\left(5-x\right)^2}{5x\left(5-x\right)}\)
\(=\dfrac{-5+x}{5x}\)
\(b,\dfrac{x+1}{x+3}+\dfrac{x-7}{x^2+x-6}+\dfrac{1}{x-2}\)
\(=\dfrac{x+1}{x+3}+\dfrac{x-7}{\left(x+3\right)\left(x-2\right)}+\dfrac{1}{x-2}\)
\(=\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}+\dfrac{x-7}{\left(x+3\right)\left(x-2\right)}+\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-2x+x-2+x-7+x+3}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2+x-6}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2+x-6}{x^2-2x+3x-6}\)
\(=\dfrac{x^2+x-6}{x^2+x-6}\)
\(=1\)