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Đặt
\(\frac{x-1}{2}\)=\(\frac{y-2}{3}\)=\(\frac{z-3}{4}\)= k
Ta có: x=2k+1
y=3k+2
z=4k+3
Theo đề ta có: 2x+3y-z=50
2(2k+1)+3(3k+2)
Xin lỗi mình giải tiếp nè, lỡ tay bấm lộn
Theo đè ta có: 2x+3y-z=50
\(\Rightarrow\) 2(2k+1)+3(3k+2}-(4z+3)=50
\(\Rightarrow\) 4k+2+9k+6-4z-3=50
\(\Rightarrow\) 9k+5=50
\(\Rightarrow\) 9k=45
\(\Rightarrow\) k=5
Thay k=5 vào, ta có: x= 2.5+1=11
y= 3.5+2=17
z=4.5+3=23
Nhớ cho mình nha
x+7/2010+x+6/2011=x+5/2012+x+4/2013
((x+7/2010)-1)+((x+6/2011)-1)=(x+5/2012)-1)+(x+4/2013)-1)
x+2017/2010+x+2017/2011-x+2017/2012-x+2017/2013=0
x+2017(1/2010+1/2011-1/2012-1/2013)=0
x+2017=0(vì 1/2010+1/2011-1/2012-1/2013<0)
x=-2017
vậy.......
tk mk nha bn
\(\left(x-\frac{1}{5}\right)^3=\left(\frac{3}{5}\right)^3\)
...............tự làm tiếp
3x+3x+2=810
3x+3x.32=81.10
3x.(1+9)=81.10
3x.10=34.10
...............tự làm tiếp
27x+4=3212
27x+4=(25)12
27x+1=260
..............tự làm tiếp
=> 2^3/2x-3/5 = 17+15 = 32 = 2^5
=> 3/2x-3/5 = 5
=> 3/2x=5+3/5 = 28/5
=> x = 27/5 : 3/2 = 56/15
k mk nha
\(\frac{2^{12}.3^5-4^6.81}{\left(2^2.3\right)^6+8^4.3^5}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\)
\(=\frac{2^{12}.\left(3^5-3^4\right)}{2^{12}.\left(3^6+3^5\right)}\)
\(=\frac{3^5-3^4}{3^6+3^5}=\frac{3^4.\left(3-1\right)}{3^5\left(3+1\right)}\)
\(=\frac{3^4.2}{3^5.4}=\frac{3^4.2}{3^4.3.4}=\frac{2}{12}=\frac{1}{6}\)
P/s: Hoq chắc ạ (: Ms lp 6 lm đại
\(\frac{x}{2}=\frac{y}{3}\)
\(\Leftrightarrow\frac{x}{8}=\frac{y}{12}\)(1)
\(\frac{y}{4}=\frac{z}{5}\)
\(\Leftrightarrow\frac{y}{12}=\frac{z}{15}\)(2)
Từ (1) (2)
\(\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)
\(\Rightarrow\hept{\begin{cases}x=2.8\\y=2.12\\z=2.15\end{cases}\Rightarrow}\hept{\begin{cases}x=16\\y=24\\z=30\end{cases}}\)
(x - 2/3)3 = -1/27
=> (x - 2/3)3 = (-1/3)3
=> x - 2/3 = -1/3
=> x = -1/3 + 2/3
=> x = 1/3
Từ bài ra ta có \(\left(x-\frac{2}{3}\right)^3=\left(\frac{-1}{3}\right)^3\)
\(\Rightarrow x-\frac{2}{3}=\frac{-1}{3}\)
\(\Rightarrow x=\frac{-1}{3}+\frac{2}{3}\)
\(\Rightarrow x=\frac{1}{3}\)
Vậy ... nếu đúng thì k nha
Ta có :
\(B=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{x}.\left(1+2+3+...+x\right)\)
\(B=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\frac{1}{x}.\frac{x.\left(x+1\right)}{2}\)
\(B=1+\frac{3}{2}+\frac{4}{2}+...+\frac{x+1}{2}\)
\(B=\frac{2+3+4+...+\left(x+1\right)}{2}\)
để B = 115 thì \(\frac{2+3+4+...+\left(x+1\right)}{2}=115\)
\(\Rightarrow\)\(\left(x+3\right)x=115.2.2\)
\(\Rightarrow\)\(\left(x+3\right)x=23.20\)
\(\Rightarrow\)x = 20
\(3-\frac{1-\frac{1}{3}}{1+\frac{1}{x}}=2\frac{2}{3}\)
\(\Rightarrow\frac{1}{3}-\frac{2}{3}:\left(1+\frac{1}{x}\right)=0\)
\(\Rightarrow\frac{1}{3}-\frac{2}{3}:\frac{x+1}{x}=0\)
\(\Rightarrow\frac{2}{3}.\frac{x}{x+1}=\frac{1}{3}\)
\(\Rightarrow\frac{2x}{3x+3}=\frac{1}{3}\)
\(\Rightarrow6x=3x+3\)
\(\Rightarrow x=1\)
x sẽ = 1