Giải các phương trình:
\(a,\dfrac{5x^2+16}{x^2-16}=\dfrac{2x-1}{x+4}-\dfrac{3x-1}{4-x}\)
\(b,\dfrac{y+1}{y-2}-\dfrac{5}{y+2}=\dfrac{12}{y^2-4}+1\)
Giải các phương trình:
\(a,\dfrac{5x^2+16}{x^2-16}=\dfrac{2x-1}{x+4}-\dfrac{3x-1}{4-x}\)
\(b,\dfrac{y+1}{y-2}-\dfrac{5}{y+2}=\dfrac{12}{y^2-4}+1\)
a)
\(\dfrac{5x^2+16}{x^2-16}=\dfrac{2x-1}{x+4}-\dfrac{3x-1}{4-x}\) (\(x\ne\pm2\))
\(\Rightarrow\dfrac{5x^2+16}{\left(x-4\right)\left(x+4\right)}-\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}-\dfrac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=0\)
\(\Rightarrow\dfrac{5x^2+16-\left(2x^2-8x-x+4\right)-\left(3x^2+12x-x-4\right)}{\left(x-4\right)\left(x+4\right)}=0\)
\(\Rightarrow\dfrac{10x+16}{x^2-16}=0\)
=> 10x + 16 =0
=> 10x = -16
=> x = \(-\dfrac{8}{5}\)
giải các phương trình sau
1, \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)
2, \(\dfrac{3}{x-5}-\dfrac{15-3x}{x^2-25}=\dfrac{3}{x+5}\)
3, \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)
1: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)
\(\Leftrightarrow\dfrac{5x^2-12}{\left(x-1\right)\left(x+1\right)}+\dfrac{3x+3}{\left(x-1\right)\left(x+1\right)}=\dfrac{5x^2-5x}{\left(x+1\right)\left(x-1\right)}\)
Suy ra: \(5x^2+3x-9=5x^2-5x\)
\(\Leftrightarrow8x=9\)
hay \(x=\dfrac{9}{8}\left(tm\right)\)
2: Ta có: \(\dfrac{3}{x-5}-\dfrac{15-3x}{x^2-25}=\dfrac{3}{x+5}\)
\(\Leftrightarrow\dfrac{3x+15}{\left(x-5\right)\left(x+5\right)}+\dfrac{3x-15}{\left(x-5\right)\left(x+5\right)}=\dfrac{3x-15}{\left(x+5\right)\left(x-5\right)}\)
Suy ra: \(6x=3x-15\)
\(\Leftrightarrow3x=-15\)
hay \(x=-5\left(loại\right)\)
2. ĐKXĐ: $x\neq \pm 5$
PT \(\Leftrightarrow \frac{3}{x-5}+\frac{3x-15}{x^2-25}=\frac{3}{x+5}\)
\(\Leftrightarrow \frac{3}{x-5}+\frac{3(x-5)}{(x-5)(x+5)}=\frac{3}{x+5}\)
\(\Leftrightarrow \frac{3}{x-5}+\frac{3}{x+5}=\frac{3}{x+5}\Leftrightarrow \frac{3}{x-5}=0\) (vô lý)
Vậy pt vô nghiệm.
3. ĐKXĐ: $x\neq \pm 4$
PT \(\Leftrightarrow \frac{-3(x+4)}{(x-4)(x+4)}-\frac{3-5x}{(x-4)(x+4)}=\frac{x-4}{(x-4)(x+4)}\)
\(\Rightarrow -3(x+4)-(3-5x)=x-4\)
\(\Leftrightarrow 2x-15=x-4\Leftrightarrow x=11\) (thỏa mãn)
Giải các phương trình sau:
a.{\(\dfrac{3x+1}{2}-\dfrac{y-2}{3}=4\)
{\(\dfrac{x-2}{3}+\dfrac{y+1}{4}=5\)
b.{(x + 5) (y - 4) = xy
{(x + 5) (y + 12) = xy
b: Ta có: \(\left\{{}\begin{matrix}\left(x+5\right)\left(y-4\right)=xy\\\left(x+5\right)\left(y+12\right)=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy-4x+5y-20-xy=0\\xy+12x+5y+60-xy=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4x+5y=20\\12x+5y=-60\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-16y=80\\-4x+5y=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-5\\-4x=20-5y=20-5\cdot\left(-5\right)=45\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-5\\x=-\dfrac{45}{4}\end{matrix}\right.\)
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 \(x=\dfrac{9}{8}\left(nhận\right)\)
Giải các hệ phương trình sau
f.{ (2x - y) (x + 3y) = 4
{ (5x + y) (x + 3y) = 24
g.{ \(\dfrac{8x-5y-3}{7}+\dfrac{11y-4x-7}{5}=12\)
{ \(\dfrac{9x+4y-13}{5}+\dfrac{3\left(x-2\right)}{4}=15\)
h.{\(\dfrac{1}{x}+\dfrac{1}{y}=2\)
{\(\dfrac{3}{x}-\dfrac{4}{y}=-1\)
h) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{3}{x}-\dfrac{4}{y}=-1\end{matrix}\right.\)\(\left(1\right)\)\(\left(đk:x,y\ne0\right)\)
Đặt \(a=\dfrac{1}{x},b=\dfrac{1}{y}\)
\(\left(1\right)\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=2\\3a-4b=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6\\3a-4b=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\7b=7\end{matrix}\right.\)\(\Leftrightarrow a=b=1\)
Thay a,b:
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{y}=1\Leftrightarrow x=y=1\left(tm\right)\)
Giải các phương trình sau:
\(a.\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\)
\(b.\dfrac{7}{x+2}=\dfrac{3}{x-5}\)
\(c.\dfrac{14}{3x-12}-\dfrac{2+x}{x-4}=\dfrac{3}{8-2x}-\dfrac{5}{6}\)
\(d.\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{x^2-1}\)
TK
https://lazi.vn/edu/exercise/giai-phuong-trinh-4x-5-x-1-2-x-x-1-7-x-2-3-x-5
a: \(\Leftrightarrow4x-5=2x-2+x\)
=>4x-5=3x-2
=>x=3(nhận)
b: =>7x-35=3x+6
=>4x=41
hay x=41/4(nhận)
c: \(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{28}{6\left(x-4\right)}-\dfrac{6\left(x+2\right)}{6\left(x-4\right)}=\dfrac{-9}{6\left(x-4\right)}-\dfrac{5\left(x-4\right)}{6\left(x-4\right)}\)
\(\Leftrightarrow28-6x-12=-9-5x+20\)
=>-6x+16=-5x+11
=>-x=-5
hay x=5(nhận)
d: \(\Leftrightarrow x^2+2x+1-\left(x^2-2x+1\right)=16\)
\(\Leftrightarrow4x=16\)
hay x=4(nhận)
1. Giải các phương trình sau:
a) \(\dfrac{7x-2}{3}=\dfrac{3x+1}{4}\) b) \(\dfrac{3x-1}{x+1}=\dfrac{2x+1}{x-1}\)
2. Tìm A : \(\dfrac{x^2+2xy+y^2}{x-y}=\dfrac{A}{x^2-y^2}\)
Giúp em với mọi người ơiii
1:
a: =>28x-8=9x+3
=>19x=11
=>x=11/19
b: =>(3x-1)(x-1)=(2x+1)(x+1)
=>3x^2-4x+1=2x^2+3x+1
=>x^2-7x=0
=>x=0 hoặc x=7
tìm x :
a) \(\dfrac{x+1}{7}+\dfrac{x+1}{8}=\dfrac{x+1}{9}+\dfrac{x+1}{10}\)
b) \(\dfrac{x+1}{12}+\dfrac{x^2}{11}=\dfrac{x+3}{10}+\dfrac{x+4}{9}\)
c) \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)
d)\(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x-3}\)
2 tìm x,y,z
a) \(\dfrac{x}{2}=\dfrac{y}{3},\dfrac{y}{4}=\dfrac{z}{5}vàx^2-y^2=-16\)
b)2x=3y,5x=7z và 3x-7y+5z=30
Giải các phương trình sau:
\(e.\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\)
\(f.\dfrac{6x+1}{x^2-7x+10}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)
\(g.\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\)
\(h.\dfrac{8}{x-8}+\dfrac{11}{x-11}=\dfrac{9}{x-9}+\dfrac{10}{x-10}\)
e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)
\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)
\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)
\(\Leftrightarrow x=-1\left(TM\right)\)