Giải pt: 30/x - 30/x+5 =1
60/x - 60/x+2 =1
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28 tháng 6 2021
1.Pt \(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=sin\left(x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\left(\dfrac{\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\\2x-\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Rightarrow x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\)\(\left(k\in Z\right)\)
2.\(sin^22x+cos^23x=1\)
\(\Leftrightarrow\dfrac{1-cos4x}{2}+\dfrac{1+cos6x}{2}=1\)
\(\Leftrightarrow cos6x=cos4x\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{5}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow x=\dfrac{k\pi}{5}\)\(\left(k\in Z\right)\) (Gộp nghiệm)
Vậy...
3. \(Pt\Leftrightarrow\left(sinx+sin3x\right)+\left(sin2x+sin4x\right)=0\)
\(\Leftrightarrow2.sin2x.cosx+2.sin3x.cosx=0\)
\(\Leftrightarrow2cosx\left(sin2x+sin3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin3x=-sin2x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\sin3x=sin\left(\pi+2x\right)\end{matrix}\right.\)(\(k\in Z\))
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\)(\(k\in Z\))\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\) (\(k\in Z\))
Vậy...
4. Pt\(\Leftrightarrow\dfrac{1-cos2x}{2}+\dfrac{1-cos4x}{2}=\dfrac{1-cos6x}{2}\)
\(\Leftrightarrow cos2x+cos4x=1+cos6x\)
\(\Leftrightarrow2cos3x.cosx=2cos^23x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\cosx=cos3x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=-k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)
Vậy...
NT
1
3 tháng 8 2019
\(\frac{60}{x}=\frac{30}{x-6}+\frac{30}{x+10}\)
\(\Leftrightarrow\frac{60}{x}=\frac{30}{x-6}+\frac{30}{x+10},Đkxđ:x\ne0,6,-10\)
\(\Leftrightarrow\frac{60}{x}-\frac{30}{x-6}-\frac{30}{x+10}=0\)
\(\Leftrightarrow\frac{60\left(x-6\right)\left(x+10\right)-30x\left(x+10\right)=30\left(x-6\right)}{x\left(x-6\right)\left(x+10\right)}\)
\(\Leftrightarrow\frac{\left(60x-360\right)\left(x+10\right)-30x^2-300x-30x^2+180x}{x\left(x-6\right)\left(x+10\right)}\)
\(\Leftrightarrow\frac{60x^2+600x-360x-3600-30x^2-300x-30x^2+180}{x\left(x-6\right)\left(x=10\right)}=0\)
\(\Leftrightarrow\frac{120x-3600}{x\left(x-6\right)\left(x+10\right)}=0\)
\(\Leftrightarrow120x-3600=0\)
\(\Leftrightarrow120x=3600\)
\(\Leftrightarrow x=30;x\ne0;x\ne6,x\ne-10\)
NK
1
7 tháng 12 2015
a/30.x=2280040
30.x=570
x=570:30=19
b/(x.30):40=840
x.30 =840.40
x.30 =33600
x =33600:30=1120
c/37200:x=60.5
37200:x=300
x=37200:300=124
d/(72000:x):60=5
72000:x =5.60
72000:x =300
x =72000:300=240
16 tháng 6 2019
\(\frac{60}{5}+x=\frac{60}{x-1}\) (x khác 1)
\(\Leftrightarrow12+x=\frac{60}{x-1}\)
\(\Leftrightarrow\frac{x^2-x-60}{x-1}=12\)\(\Leftrightarrow x^2-x-60=12x-12\)
\(\Leftrightarrow x^2+11x-48=0\)
ĐỂ PT CÓ NGHIỆM THÌ
\(\Delta=11^2+4.48=121+192=323>0\)
giải tiếp là ra
9 tháng 8 2023
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)
18 tháng 9 2016
a, 245 - 5 . ( 16 + x ) = 140
5 . ( 16 + x ) = 245 - 140
5 . ( 16 + x ) = 145
16 + x = 145 : 5
16 + x = 29
x = 29 - 16
x = 13 .
b, ( x - 1945 ) . 5 = 50
x - 1945 = 50 : 5
x - 1945 = 10
x = 10 + 1945
x = 1955 .
c, 30 . ( 60 - x ) = 30
60 - x = 30 : 30
60 - x = 1
x = 60 - 1
x = 59 .
d, [ ( 250 - 25 ) : 15 ] : x = ( 450 - 60 ) : 130
[ 225 : 15 ] : x = 390 : 130
15 : x = 3
x = 15 : 3
x = 5 .
e, x : [ ( 1800 + 600 ) ] = 560 : ( 315 - 35 )
x : 2400 = 560 : 280
x : 2400 = 2
x = 2 . 2400
x = 4800 .
f, x . ( x + 1 ) = 2 + 4 + 6 + 8 + 10 + ... + 2500
2 + 4 + 6 + 8 + 10 + ... + 2500
Số số hạng của dãy số trên là :
( 2500 - 2 ) : 2 + 1 = 1250 ( số hạng )
=> 2 + 4 + 6 + 8 + 10 + 2500
= ( 2 + 2500 ) . 1250 : 2
= 2502 . 1250 : 2
= 3127500 : 2
= 1563750 .
Ta có :
x . ( x + 1 ) = 1563750
Mà : 1563750 = 1250 . 1251
=> x = 1250 .
CM
8 tháng 3 2018
37 - 5 x 5 = 12 : Đ
180 : 6 + 30 = 60 : Đ
30 + 60 x 2 = 150 : Đ
282 - 100 : 2 = 91 : S
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 }