60/5+x=60/x-1 giải pt
Giải pt: 30/x - 30/x+5 =1
60/x - 60/x+2 =1
Bài làm:
1) đk: \(x\ne0;x\ne-5\)
Ta có: \(\frac{30}{x}-\frac{30}{x+5}=1\)
\(\Leftrightarrow\frac{30\left(x+5\right)-30x}{x\left(x+5\right)}=1\)
\(\Leftrightarrow x^2+5x=150\)
\(\Leftrightarrow x^2+5x-150=0\)
\(\Leftrightarrow\left(x-10\right)\left(x+15\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-10=0\\x+15=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=10\\x=-15\end{cases}}\)
2) đk: \(x\ne0;x\ne-2\)
Ta có: \(\frac{60}{x}-\frac{60}{x+2}=1\)
\(\Leftrightarrow\frac{60\left(x+2\right)-60x}{x\left(x+2\right)}=1\)
\(\Leftrightarrow x^2+2x=120\)
\(\Leftrightarrow x^2+2x-120=0\)
\(\Leftrightarrow\left(x-10\right)\left(x+12\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-10=0\\x+12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=10\\x=-12\end{cases}}\)
\(\frac{30}{x}-\frac{30}{x+5}=1\)( ĐKXĐ : \(x\ne0;x\ne-5\))
<=> \(30\left(\frac{1}{x}-\frac{1}{x+5}\right)=1\)
<=> \(30\left(\frac{x+5}{x\left(x+5\right)}-\frac{x}{x\left(x+5\right)}\right)=1\)
<=> \(30\left(\frac{5}{x\left(x+5\right)}\right)=1\)
<=> \(\frac{5}{x\left(x+5\right)}=\frac{1}{30}\)
<=> \(5\cdot30=x\left(x+5\right)\)
<=> \(x^2+5x-150=0\)
<=> \(x^2+15x-10x-150=0\)
<=> \(x\left(x+15\right)-10\left(x+15\right)=0\)
<=> \(\left(x-10\right)\left(x+15\right)=0\)
<=> \(\orbr{\begin{cases}x-10=0\\x+15=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-15\end{cases}}\)( tmđk )
Vậy S = { 10 ; -15 }
\(\frac{60}{x}-\frac{60}{x+2}=1\)( ĐKXĐ : \(x\ne0;x\ne-2\))
<=> \(60\left(\frac{1}{x}-\frac{1}{x+2}\right)=1\)
<=> \(60\left(\frac{x+2}{x\left(x+2\right)}-\frac{x}{x\left(x+2\right)}\right)=1\)
<=> \(60\left(\frac{2}{x\left(x+2\right)}\right)=1\)
<=> \(\frac{2}{x\left(x+2\right)}=\frac{1}{60}\)
<=> \(2\cdot60=x\left(x+2\right)\)
<=> \(x^2+2x-120=0\)
<=> \(x^2+12x-10x-120=0\)
<=> \(x\left(x+12\right)-10\left(x+12\right)=0\)
<=> \(\left(x-10\right)\left(x+12\right)=0\)
<=> \(\orbr{\begin{cases}x-10=0\\x+12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-12\end{cases}}\)
Vậy S = { 10 ; -12 }
a, \(\frac{30}{x}-\frac{30}{x+5}=1\) b, \(\frac{60}{x}-\frac{60}{x+2}=1\)
ĐKXĐ: x khác 0 ĐKXĐ: x khác 0
x khác -5 x khác -2
Ta có: Ta có:
\(\frac{30}{x}-\frac{30}{x+5}=1\) \(\frac{60}{x}-\frac{60}{x+2}=1\)
<=>\(\frac{30\left(x+5\right)}{x\left(x+5\right)}-\frac{30x}{x\left(x+5\right)}=\frac{x\left(x+5\right)}{x\left(x+5\right)}\) <=>\(\frac{60\left(x+2\right)}{x\left(x+2\right)}-\frac{60x}{x\left(x+2\right)}=\frac{x\left(x+2\right)}{x\left(x+2\right)}\)
=>30(x+5)-30x=x(x+5) =>60(x+2)-60x=x(x+2)
<=>30x+150-30x=x2+5x <=>60x+120-60x=x2+2x
<=>150=x2+5x <=>120=x2+2x
<=>0=x2+5x-150 <=>0=x2+2x-120
<=>0=x2-10x+15x-150 <=>x2-10x+12x-120
<=>0=(x2-10x)+(15x-150) <=>(x2-10x)+(12x-120)
<=>0=x(x-10)+15(x-10) <=>x(x-10)+12(x-10)
<=>0=(x-10)(x+15) <=>(x-10)(x+12)
<=>x-10=0 hoặc x+15=0 <=>x-10=0 hoặc x+12=0
1, x-10=0 2,x+15=0 1, x-10=0 2, x+12=0
<=>x=10 <=>x=-15 <=>x=10 <=>x=-12
( thỏa mãn ĐKXĐ) (thỏa mãn ĐKXĐ)
Vậy tâp nghiệm của PT là S={10;-15} Vậy tập ngiệm của PT là S={10;-12}
giải pt sau
a)\(\dfrac{60}{x}=\dfrac{4}{3}+\dfrac{60-x}{x+4}\)
b)\(\dfrac{100}{x}-\dfrac{100}{x+20}=\dfrac{5}{6}\)
c)\(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)
Helppppp
b: \(\Leftrightarrow\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\)
=>\(\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\)
=>x*(x+20)=400*6=2400
=>x^2+20x-2400=0
=>(x+60)(x-40)=0
=>x=-60 hoặc x=40
c: \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)
=>(2x+1)^2-(2x-1)^2=8
=>4x^2+4x+1-4x^2+4x-1=8
=>8x=8
=>x=1(nhận)
giải pt
x/60+x/45=7
\(\dfrac{x}{60}+\dfrac{x}{45}=7\)
\(\Leftrightarrow\dfrac{315x+420x-132300}{18900}=0\)
\(\Leftrightarrow735x=132300\)
\(\Leftrightarrow x=180\)
giải pt
\(\dfrac{1}{x}+\dfrac{1}{x+50}=\dfrac{1}{60}\)
\(\dfrac{1}{x}+\dfrac{1}{x+50}=\dfrac{1}{60}\left(x\ne0;x\ne-5\right)\)
\(pt\Leftrightarrow\dfrac{x+50}{x\left(x+50\right)}+\dfrac{x}{x\left(x+50\right)}=\dfrac{1}{60}\)
\(\Leftrightarrow\dfrac{2x+50}{x\left(x+50\right)}=\dfrac{1}{60}\Leftrightarrow x\left(x+50\right)=60\left(2x+50\right)\)
\(\Leftrightarrow x^2+50x=120x+3000\)
\(\Leftrightarrow x^2-70x-3000=0\)
\(\Leftrightarrow x^2-100x+30x-3000=0\)
\(\Leftrightarrow x\left(x-100\right)+30\left(x-100\right)=0\)
\(\Leftrightarrow\left(x+30\right)\left(x-100\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+30=0\\x-100=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-30\\x=100\end{matrix}\right.\)
Giải hệ pt sau
4/x +5/y = 2/3
5/x+4/y = 41/60
Mk cần gấp các bn giúpmk nhé
\(\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{5}{y}=\dfrac{2}{3}\\\dfrac{5}{x}+\dfrac{4}{y}=\dfrac{41}{60}\end{matrix}\right.\left(I\right)\)
Đặt \(:\left\{{}\begin{matrix}t=\dfrac{1}{x}\\u=\dfrac{1}{y}\end{matrix}\right.\)
\(\left(I\right):\left\{{}\begin{matrix}4t+5u=\dfrac{2}{3}\\5t+4u=\dfrac{41}{60}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}20t+25u=\dfrac{10}{3}\\20t+16u=\dfrac{41}{15}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9u=\dfrac{3}{5}\\20t+16u=\dfrac{41}{15}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u=\dfrac{1}{15}\\t=\dfrac{1}{12}\end{matrix}\right.\)
Với \(:\left\{{}\begin{matrix}t=\dfrac{1}{12}\\u=\dfrac{1}{15}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{12}\\\dfrac{1}{y}=\dfrac{1}{15}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=12\\y=15\end{matrix}\right.\)
Vậy nghiệm hệ phương trình là \(\left(12;15\right)\)
Giải pt sau: (Lớp 11)
sin(x+60°) - cos2x = 0
\(x=10\)
mk ấn máy tính thì nó bằng thế
mk mới lp 9 thôi mà
giai PT
\(\dfrac{x}{60}+ \dfrac{x}{45}+1,5=\dfrac{32}{5}\)
x/60 + x/45 + 1,5 = 32/5
3x + 4x + 270 = 1152
7x + 270 = 1152
7x = 1152 − 270
7x = 882
x = 882/7
x = 126
\(\dfrac{x}{60}+\dfrac{x}{45}+1,5=\dfrac{32}{5}\)
\(\Leftrightarrow\)\(\dfrac{x}{60}+\dfrac{x}{45}+\dfrac{3}{2}=\dfrac{32}{5}\)
\(\Leftrightarrow\dfrac{3x}{180}+\dfrac{4x}{180}+\dfrac{270}{180}=\dfrac{1152}{180}\)
\(\Leftrightarrow7x+270=1152\)
\(\Leftrightarrow7x=882\)
\(\Leftrightarrow x=126\)
giải pt theo cách giải \(\Delta\)
a,\(\dfrac{60}{x}-\dfrac{1}{2}=\dfrac{60}{x+2}\)
b,\(\dfrac{x+2}{x-1}=\dfrac{4x^2-11x+2}{\left(x+2\right)\left(1-x\right)}\)
c,\(\left(3x^2+2\right)^2-15x^3-10x=0\)
b: \(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-1\right)\left(x+2\right)}=\dfrac{-4x^2+11x-2}{\left(x+2\right)\left(x-1\right)}\)
\(\Leftrightarrow x^2+4x+4+4x^2-11x+2=0\)
\(\Leftrightarrow5x^2-7x+6=0\)
hay \(x\in\varnothing\)
c: \(\Leftrightarrow\left(3x^2+2\right)^2-5x\left(3x^2+2\right)=0\)
=>3x^2-5x+2=0
=>3x^2-3x-2x+2=0
=>(x-1)(3x-2)=0
=>x=2/3 hoặc x=1
Giải pt
\(\frac{60}{x}=\frac{30}{x-6}+\frac{30}{x+10}\)
\(\frac{60}{x}=\frac{30}{x-6}+\frac{30}{x+10}\)
\(\Leftrightarrow\frac{60}{x}=\frac{30}{x-6}+\frac{30}{x+10},Đkxđ:x\ne0,6,-10\)
\(\Leftrightarrow\frac{60}{x}-\frac{30}{x-6}-\frac{30}{x+10}=0\)
\(\Leftrightarrow\frac{60\left(x-6\right)\left(x+10\right)-30x\left(x+10\right)=30\left(x-6\right)}{x\left(x-6\right)\left(x+10\right)}\)
\(\Leftrightarrow\frac{\left(60x-360\right)\left(x+10\right)-30x^2-300x-30x^2+180x}{x\left(x-6\right)\left(x+10\right)}\)
\(\Leftrightarrow\frac{60x^2+600x-360x-3600-30x^2-300x-30x^2+180}{x\left(x-6\right)\left(x=10\right)}=0\)
\(\Leftrightarrow\frac{120x-3600}{x\left(x-6\right)\left(x+10\right)}=0\)
\(\Leftrightarrow120x-3600=0\)
\(\Leftrightarrow120x=3600\)
\(\Leftrightarrow x=30;x\ne0;x\ne6,x\ne-10\)