giải pt
x/60+x/45=7
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 phương trình x>60+x>45+7
60/5+x=60/x-1 giải pt
\(\frac{60}{5}+x=\frac{60}{x-1}\) (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
Giải PT:
a,\(\sqrt[3]{x+45}-\sqrt[3]{x-16}=1\)
b.\(\sqrt{x+1}-\sqrt{x-7}=\sqrt{12-x}\)
Giải PT x/40-x/45=3/2
\(\frac{x}{40}-\frac{x}{45}=\frac{3}{2}\)
\(\Leftrightarrow\frac{9x-8x}{360}=\frac{3}{2}\)
\(\Leftrightarrow2x=3.360\)
\(\Leftrightarrow2x=1080\)
\(\Leftrightarrow x=540\)
\(\frac{x}{40}-\frac{x}{45}=\frac{3}{2}\)
\(\Leftrightarrow\frac{18x}{720}-\frac{16x}{720}=\frac{1080}{720}\)
\(\Rightarrow18x-16x=1080\)(KHỬ MẪU)
\(\Leftrightarrow2x=1080\)
\(\Leftrightarrow x=\frac{1080}{2}\)
\(\Leftrightarrow x=540\)
Vậy tập nghiệm của phương trình là \(S=\left\{540\right\}\)
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:
X-45/55 + X-47/53 = X-55/45 +X-53/47
các bn giúp mik vs
<=>x-45/55 -1 + x-47/53 -1=x-55/45 -1 + x-53/47-1
<=>x-100/55 + x-100/53 = x-100/45 + x-100/47
<=>(x-100)(1/55 + 1/53 - 1/45 - 1/47 )=0
vi (1/55 + 1/53 - 1/45 - 1/47 ) luon khac 0 nen x-100=0 <=>x=100
giải pt :
x3-x2-24x+45=0
Giải pt : \({x\over 30 } - {x\over 45 } = {1\over 4 }\)