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a) x-12=-9+15
x-12=24
x=24+12
x=36
Vậy x=36
b)4x -12=400
4x=412
x=412:4
x=103
Vậy x=103
c)2x -35=15
2x=50
x=50:2
x=15
Vậy x=15
d)3x+17=2
3x=-15
x=(-15):3
x=-5
Vậy x=-5
e)\(\frac{-5}{8}=\frac{x}{16}\)
\(\Rightarrow\left(-5\right).16=8.x\)
\(\Rightarrow8x=-80\)
\(\Rightarrow x=-10\)
f)\(\frac{y}{10}=-\frac{4}{8}\)
\(\Rightarrow8y=\left(-4\right).10\)
\(\Rightarrow8y=-40\)
\(\Rightarrow y=-5\)
a) x - 12 = -9 + 15
=> x - 12 = 6
=> x = 6 + 12 = 18
b) 4x - 12 = 400
=> 4x = 400 + 12 = 412
=> x = 412 : 4 = 103
c) 2x - 35 = 15
=> 2x = 15 + 35 = 50
=> x = 50 : 2 = 25
d) 3x + 17 = 2
=> 3x = 2 - 17
=> 3x = -15
=> x = -15 : 3 = -5
e) \(\frac{-5}{8}=\frac{x}{16}\)
\(=>\frac{-10}{16}=\frac{x}{16}\)
=> x = -10
f) \(\frac{y}{10}=\frac{-4}{8}\)
\(=>\frac{y}{10}=\frac{-1}{2}\)
\(=>\frac{y}{10}=\frac{-5}{10}\)
=> y = -5
a,\(\frac{11}{12}-\left(\frac{5}{42}-x\right)=\frac{15}{28}-\frac{11}{12}\)
\(\Leftrightarrow\frac{11}{12}-\frac{5}{42}+x=\frac{15}{28}-\frac{11}{12}\)
\(\Leftrightarrow x=\frac{15}{28}-\frac{11}{12}-\frac{11}{12}+\frac{5}{42}\)
\(\Leftrightarrow x=\left(\frac{15}{28}+\frac{5}{42}\right)-\left(\frac{11}{12}+\frac{11}{12}\right)\)
\(\Leftrightarrow x=\frac{55}{84}-\frac{11}{6}\)
\(\Leftrightarrow x=\frac{-33}{28}\)
b, \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
a. 2(x-1) + (x+2) - (x+3) = 15 - (x+1)
=>2x-2+x+2-x-3=15-x-1
=>(2x+x-x)-2+2-3=15-1-x
=>2x-3=14-x
=>3x=17
=>x=17/3
b. x+1/15 + x+2/14 = x+4/12 + x+5/11
\(\Rightarrow\frac{x+1}{15}+1+\frac{x+2}{14}+1=\frac{x+4}{12}+1+\frac{x+5}{11}+1\)
\(\Rightarrow\frac{x+16}{15}+\frac{x+16}{14}=\frac{x+16}{12}+\frac{x+16}{11}\)
\(\Rightarrow\frac{x+16}{15}+\frac{x+16}{14}-\frac{x+16}{12}-\frac{x+16}{11}=0\)
\(\Rightarrow\left(x+16\right)\left(\frac{1}{15}+\frac{1}{14}-\frac{1}{12}-\frac{1}{11}\right)=0\)
\(\Rightarrow x+16=0\).Do \(\frac{1}{15}+\frac{1}{14}-\frac{1}{12}-\frac{1}{11}\ne0\)
=>x=-16
\(\frac{15}{x-9}=\frac{12}{y-12}=\frac{40}{z-24}\)
=> \(\frac{x-9}{15}=\frac{y-12}{12}=\frac{z-24}{40}\)
Đặt \(\frac{x-9}{15}=\frac{y-12}{12}=\frac{z-24}{40}=k\Rightarrow\hept{\begin{cases}x-9=15k\\y-12=12k\\z-24=40k\end{cases}}\)
=> \(\hept{\begin{cases}x=15k+9\\y=12k+12\\z=40k+24\end{cases}}\)
Mà xy = 200
=> \(\left(15k+9\right)\left(12k+12\right)=200\)
=> 15(12k + 12) + 9(12k + 12) = 200
=> 180k + 180 + 108k + 108 = 200
=> 288k + 216 = 200
=> 288k = -16
Đề của bạn chắc chắn đúng chứ , mình thấy sai rồi đấy :v
Do x = 11 => x + 1 = 12
Ta có :
D = \(x^{17}-\left(x+1\right)x^{16}+\left(x+1\right).x^{15}+...+\left(x+1\right).1\)
= \(x^{17}-x^{17}-x^{16}+x^{16}+x^{15}+...+x+1\)
= \(0+0+0+...+1\)
= \(1\)
Vậy D = 1
Tham khảo nha !!! Chúc học tốt !!!
Ta có : \(\frac{x}{12}=\frac{y}{15}\)và \(y+x=27\)
Áp dụng tính chất của dãy tỉ số bằng nhau , ta có :
\(\frac{x}{12}=\frac{y}{15}=\frac{x+y}{12+15}=\frac{27}{27}=1\)
\(\Rightarrow\frac{x}{12}=1\Rightarrow x=12\)
\(\frac{y}{15}=1\Rightarrow y=15\)
Vậy x = 12 ; y = 15
\(U\left(15\right)=\left\{1;3;5;15\right\}\)
mà \(12\ge x\le15\)
\(\Rightarrow x\in\left\{1;3;5\right\}\)
Đề bài yêu cầu gì bạn ?