Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\Leftrightarrow\frac{x-1}{117}+1+\frac{x-2}{118}+1+\frac{x-3}{119}=\frac{x-4}{120}+1+\frac{x-5}{121}+1+\frac{x-6}{122}+1\)
\(\Leftrightarrow\frac{x+116}{117}+\frac{x+116}{118}+\frac{x+116}{119}-\frac{x+116}{120}-\frac{x+116}{121}-\frac{x+116}{122}=0\)
\(\Leftrightarrow\left(x+116\right)\left(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\right)=0\)
\(\Leftrightarrow x+116=0\Leftrightarrow x=-116\)
\(\frac{x-1}{117}+\frac{x-2}{118}+\frac{x-3}{119}=\frac{x-4}{120}+\frac{x-5}{121}+\frac{x-6}{122}\)
\(\Leftrightarrow\frac{x-1}{117}+1+\frac{x-2}{118}+1+\frac{x-3}{119}+1=\frac{x-4}{120}+1+\frac{x-5}{121}+1+\frac{x-6}{122}+1\)
\(\Leftrightarrow\frac{x+116}{117}+\frac{x+116}{118}+\frac{x+116}{119}-\frac{x+116}{120}-\frac{x+116}{121}-\frac{x+116}{122}=0\)
\(\Leftrightarrow\left(x+116\right)\left(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\right)=0\)
Vì \(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\ne0\)
Nên x + 116 = 0
<=> x = -116
Bài 1: Tìm x, y, z
\(\frac{x}{3}=\frac{y}{4}=>\frac{x}{3\times3}=\frac{y}{4\times3}=>\frac{x}{9}=\frac{y}{12}\)
\(\frac{y}{3}=\frac{z}{5}=>\frac{y}{3.4}=\frac{z}{5.4}=>\frac{y}{12}=\frac{z}{20}\)
=> \(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\)
- Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\) -> \(\frac{2x}{2\times9}=\frac{3y}{3\times12}=\frac{z}{20}\) -> \(\frac{2x}{18}=\frac{3y}{36}=\frac{z}{20}\)
-> \(\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)
\(\frac{x}{9}=3\rightarrow x=27\)
\(\frac{y}{12}=3\rightarrow y=36\)
\(\frac{z}{20}=3\rightarrow z=60\)
Vậy x = 27 ; y = 36 ; z = 60
Bài 2 : Tìm x, y:
5x = 2y và x.y = 40
Vì 5x = 2y => \(\frac{x}{2}=\frac{y}{5}\)
Cách 1:
\(\frac{x}{2}=\frac{y}{5}\) và x.y = 40
Đặt \(\frac{x}{2}=\frac{y}{5}\) = k
=> x = 2.k ; y = 5.k
x.y = 40 -> 2k = 5k = 40
-> 10 . \(k^2\) = 40
-> \(k^2\) = 4 -> k = 2 hoặc k = -2
k = 4 ta có : \(\frac{x}{2}=\frac{y}{5}=2->x=4;y=10\)
k = -4 ta có : \(\frac{x}{2}=\frac{y}{5}=-2->x=-4;y=-10\)
Cách 2:
\(\frac{x}{2}=\frac{y}{5}->\frac{x.x}{2}=\frac{x.y}{5}->\frac{x^2}{2}=\frac{40}{5}=\frac{x^2}{2}=8\)
=> \(x^2\) = 8 . 2 = 16 -> x = 4 hoặc -4
x = 4 -> 4.y = 40 => y = 10
x = -4 -> (-4).y = 40 => y = -10
Vậy x = 4 hoặc -4
y = 10 hoặc -10
\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12}\left(1\right)\\\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{15}\left(2\right)\)
Từ (1),(2) suy ra \(\frac{x}{9}=\frac{y}{12}=\frac{z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x}{9}=\frac{y}{12}=\frac{z}{15}=\frac{2x}{18}=\frac{-3y}{-36}=\frac{z}{15}=\frac{2x-3y+z}{18-\left(-36\right)+15}=\frac{6}{69}=\frac{2}{23}\)Suy ra x =\(\frac{2}{23}\cdot9=\frac{18}{23}\)
\(y=\frac{2}{23}\cdot12=\frac{24}{23}\\ z=\frac{2}{23}.15=\frac{30}{23}\)
\(\Rightarrow\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)=0\)
\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}=0\)
\(\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\right)=0\)
MÀ\(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\ne0\)
NÊN \(x+329=0\)
\(\Rightarrow x=-329\)
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}=-4\)
\(\Rightarrow\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1=0\)
\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}=0\)
\(\Rightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\right)=0\)
Vì \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\ne0\Rightarrow x+329=0\Rightarrow x=-329\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có
\(\frac{x+2+x+3+x+4+x+5}{327+326+325+324}=-4\)
=>\(\frac{4x+14}{1302}=-4\)
4x+14=(-4).1302
4x+14=-5208
4x=(-5208)-14
4x=-5222
x=-1305,5
a ) \(\frac{x}{6}+\frac{x}{4}=\frac{5}{7}\)
\(\Leftrightarrow x\left(\frac{1}{6}+\frac{1}{4}\right)=\frac{5}{7}\)
\(\Leftrightarrow\frac{5}{12}x=\frac{5}{7}\)
\(\Rightarrow x=\frac{5}{7}:\frac{5}{12}\)
\(\Rightarrow x=\frac{12}{7}\)
b ) Nếu \(xy=5\) thì :
\(M=x^2y-xy^2-xy.x+xy.y-12\)
\(=x^2y-xy^2-x^2y+xy^2-12\)
\(=\left(xy^2-x^2y\right)+\left(-xy^2+xy^2\right)-12\)
\(=-12\)
\(\left|x-\frac{5}{4}\right|-\left|x+\frac{2}{3}\right|=0\)
\(\left|x-\frac{5}{4}\right|=\left|x+\frac{2}{3}\right|\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{5}{4}=x+\frac{2}{3}\\x-\frac{5}{4}=-\frac{2}{3}-x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}0x=\frac{23}{12}\\2x=\frac{7}{12}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\text{không có giá trị x nào thỏa mãn}\\x=\frac{7}{24}\end{cases}}\)
\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)\right|+\frac{2}{5}\)
\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=3,2+\frac{2}{5}\)
\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\frac{18}{5}\)
\(\left|x-\frac{1}{3}\right|=\frac{18}{5}-\frac{4}{5}\)
\(\left|x-\frac{1}{3}\right|=\frac{14}{5}\)
TH1: \(x-\frac{1}{3}=\frac{14}{5}\)
\(x=\frac{14}{5}+\frac{1}{3}\)
\(x=\frac{47}{15}\)
TH2: \(x-\frac{1}{3}=\frac{-14}{5}\)
\(x=\frac{-14}{5}+\frac{1}{3}\)
\(x=\frac{-37}{15}\)
KL: x = 47/15 hoặc x = -37/15
\(\frac{x+4}{5}=\frac{20}{x+4}\)
\(\Leftrightarrow\left(x+4\right)^2=100\)
\(\Leftrightarrow x+4=\pm\sqrt{100}\)
\(\Leftrightarrow\orbr{\begin{cases}x+4=10\\x+4=-10\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=-14\end{cases}}\)
x=6 nha