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Giải
\(\frac{x+50}{x+68}=\frac{x-15}{x-10}\)
Theo định nghĩa hai phân thức bằng nhau, ta có:\(\left(x+50\right)\left(x-10\right)=\left(x+68\right)\left(x-15\right)\)
\(\Leftrightarrow x\left(x+50\right)-10\left(x+50\right)=x\left(x+68\right)-15\left(x+68\right)\)
\(\Leftrightarrow x^2+50x-10x-500=x^2+68x-15x-1020\)
\(\Leftrightarrow x^2+40x-500=x^2+53x-1020\)
\(\Leftrightarrow13x-520=0\)
\(\Leftrightarrow13x=520+0\)
\(\Leftrightarrow13x=520\)
\(\Leftrightarrow x=\frac{520}{13}\)
\(\Leftrightarrow x=40\)
Vậy nghiệm duy nhất của phương trình là 40
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x(x - 15) = -50
x² - 15x + 50 = 0
x² - 10x - 5x + 50 = 0
(x² - 10x) - (5x - 50) = 0
x(x - 10) - 5(x - 10) = 0
(x - 10)(x - 5) = 0
x - 10 = 0 hoặc x - 5 = 0
*) x - 10 = 0
x = 0 + 10
x = 10
*) x - 5 = 0
x = 0 + 5
x = 5
Vậy x = 10; x = 5
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`a,`\(2^x -15= 2^4+1\)
`-> 2^x-15=17`
`-> 2^x=17+15`
`-> 2^x=32`
`-> 2^x=2^5`
`-> x=5`
`b,` Có phải đề là \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\) ?
`=>`\(\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}+1\)
`=>`\(\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}-\dfrac{x+3+63}{63}-\dfrac{x+4+62}{62}=0\)
`=>`\(\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)
`=>`\(\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)
Mà `1/65+1/64-1/63-1/62 \ne 0`
`-> x+66=0`
`-> x=-66`
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a: =>2^x=2^4+16=32
=>x=5
b: Sửa đề: \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)
=>\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+2}{64}+1\right)=\left(\dfrac{x+3}{63}+1\right)+\left(\dfrac{x+4}{62}+1\right)\)
=>x+66=0
=>x=-66
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\(\frac{x}{2}=\frac{y}{3};\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)
Theo tính chất dãy tỉ số bằng nhau
\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\Rightarrow x=16;y=24;z=30\)
x/2 = y/3, y/4 = x/5 và x+y-z=10
ta có:
x/2 = y3
⇒ x/8 = y/12 (1)
y/4 = z/5
⇒ y/12 = z/15 (2)
từ (1) và (2)
⇒ x/8 = y/12 = z/15
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
x/8 = y/13 = z/15 = x+y-z/8+13-15 = 10/5 = 2
x/8 = 2 ⇒ x = 8 . 2 = 16
y/13 = 2 ⇒ y = 13 . 2 = 24
z/15 = 2 ⇒ z = 15 . 2 = 30
vậy x = 16
y = 24
z = 30
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\(\hept{\begin{cases}x\left(x-y\right)=\frac{3}{10}\\y\left(x-y\right)=-\frac{3}{50}\end{cases}}\Rightarrow\frac{x}{y}=-5\Rightarrow x=-5y\)
\(x\left(x-y\right)=-5y\left(-5y-y\right)=30y^2=\frac{3}{10}\Leftrightarrow\orbr{\begin{cases}y=\frac{1}{10}\Rightarrow x=-\frac{1}{2}\\y=-\frac{1}{10}\Rightarrow x=\frac{1}{2}\end{cases}}\)
\(\frac{x+50}{x+61}=\frac{x-15}{x-10}\)
\(\Leftrightarrow\left(x+50\right).\left(x-10\right)=\left(x-15\right).\left(x+61\right)\)
\(\Rightarrow x^2-10x+50x-500=x^2+61x-15x-915\)
\(\Rightarrow40x-500=46x-915\)
\(\Rightarrow415=6x\Rightarrow x=\frac{415}{6}\)