Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Theo bài ra ta có : \(x+y+z=49\)
\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Rightarrow\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được :
\(\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}\\ =\dfrac{12x+12y+12z}{18+16+15}\\ =\dfrac{12\left(x+y+z\right)}{49}\\ =\dfrac{12\cdot49}{49}\\ =12\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{12x}{18}=12\Rightarrow12x=216\Rightarrow x=18\\\dfrac{12y}{16}=12\Rightarrow12y=192\Rightarrow y=16\\\dfrac{12z}{15}=12\Rightarrow12z=180\Rightarrow z=15\end{matrix}\right.\)
\(\text{Vậy }x=18\\ y=16\\ z=15\)
b) Theo bài ra ta có : \(2x+3y-z=50\)
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\\ \Rightarrow\dfrac{2\left(x-1\right)}{4}=\dfrac{3\left(y-2\right)}{9}=\dfrac{z-3}{4}\\ \Rightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được :
\(\dfrac{2x-2}{4}=\dfrac{3y-2}{9}=\dfrac{z-3}{4}=\\ \dfrac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+9-4}\\ =\dfrac{2x-2+3y-6-z+3}{9}\\ =\dfrac{\left(2x+3y-z\right)-\left(2+6-3\right)}{9}\\ =\dfrac{50-5}{9}\\ =\dfrac{45}{9}\\ =5\\ \)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2x-2}{4}=5\Rightarrow2x-2=20\Rightarrow2x=22\Rightarrow x=11\\\dfrac{3y-6}{9}=5\Rightarrow3y-6=45\Rightarrow3y=51\Rightarrow y=17\\\dfrac{z-3}{4}=5\Rightarrow z-3=20\Rightarrow z=23\end{matrix}\right.\)
\(\text{Vậy }x=11\\ y=17\\ z=23\)
làm câu b , bài 1 nhé
A =(ghi lại )
=> 2A=2+22+23+24+....+2100+2101
=> 2A - A = A = 2+22+23+24+....+2100+2101 -1 -2-22-23-....-2100
=>A = 2101-1 < 2101
Vậy A < B
Bài 1:
a) \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{19}{81.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)
b) Ta có: \(A=2^0+2^1+...+2^{100}\)
\(\Rightarrow2A=2+2^2+...+2^{101}\)
\(\Rightarrow2A-A=2^{101}-2^0\)
\(\Rightarrow A=2^{201}-1< 2^{101}\)
\(\Rightarrow A< B\)
Vậy A < B
b \(\Leftrightarrow3^x\cdot9+4\cdot3^x\cdot3+3^x\cdot\dfrac{1}{3}=6^6\)
\(\Leftrightarrow3^x=6^6:\left(9+4\cdot3+\dfrac{1}{3}\right)=2187\)
hay x=7
c: \(\Leftrightarrow2^{x-1}=24-16+3-3=8\)
=>x-1=3
hay x=4
d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{-3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{-2x+7y-3z}{6+28-15}=\dfrac{171}{19}=9\)
Do đó: x=-27; y=36; z=45
a) \(\dfrac{49}{81}=\dfrac{7^x}{9^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{7}{9}\right)^2=\left(\dfrac{7}{9}\right)^x\)\(\Rightarrow x=2\)
b) \(\dfrac{-64}{343}=\left(-\dfrac{4^x}{7^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{4}{7}\right)^3=\left(-\dfrac{4}{7}\right)^x\) \(\Rightarrow x=3\)
c) \(\dfrac{9}{144}=\dfrac{3^x}{12^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{3}{12}\right)^2=\left(\dfrac{3}{12}\right)^x\Rightarrow x=2\)
d) \(-\dfrac{1}{32}=\left(-\dfrac{1^x}{2^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^5=\left(-\dfrac{1}{2}\right)^x\Rightarrow x=5\)
Mong bạn xem lại đề bài.
a, \(\left|3x-4\right|+\left|3y+5\right|=0\)
Ta có :
\(\left|3x-4\right|\ge0\forall x;\left|3y+5\right|\ge0\forall x\\ \)
\(\Rightarrow\left|3x-4\right|+\left|3y+5\right|\ge0\forall x\\ \Rightarrow\left\{{}\begin{matrix}3x-4=0\\3y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=4\\3y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=-\dfrac{5}{3}\end{matrix}\right.\\ Vậy.........\)
b, \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\)
Ta có :
\(\left|x+\dfrac{19}{5}\right|\ge0\forall x;\left|y+\dfrac{1890}{1975}\right|\ge0\forall y;\left|z-2004\right|\ge0\forall z \)
\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{19}{5}=0\\y+\dfrac{1890}{1975}=0\\z-2004=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{19}{5}\\y=-\dfrac{1890}{1975}\\z=2004\end{matrix}\right.\\ Vậy............\)
c, \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\)
Ta có : \(\left|x+\dfrac{9}{2}\right|\ge0\forall x;\left|y+\dfrac{4}{3}\right|\ge0\forall y;\left|z+\dfrac{7}{2}\right|\ge0\forall z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{2}\\y=-\dfrac{4}{3}\\z=-\dfrac{7}{2}\end{matrix}\right.\\ Vậy............\)
d, \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)
Ta có :
\(\left|x+\dfrac{3}{4}\right|\ge0\forall x;\left|y-\dfrac{1}{5}\right|\ge0\forall y;\left|x+y+z\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\z=0-\dfrac{1}{5}+\dfrac{3}{4}=\dfrac{11}{20}\end{matrix}\right.\\ Vậy.......\)
e, Câu cuối bn làm tương tự như câu a, b, c nhé!
a, \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Leftrightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy x = -1
b, \(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2014}+1\right)+\left(\dfrac{x+3}{2015}+1\right)=\left(\dfrac{x+2}{2016}+1\right)+\left(\dfrac{x+1}{2017}+1\right)\)\(\Leftrightarrow\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
\(\Leftrightarrow\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}-\dfrac{x+2018}{2016}-\dfrac{x+2018}{2017}=0\)
\(\Leftrightarrow\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\)
\(\Leftrightarrow xx+2018=0\Leftrightarrow x=-2018\)
Vậy x = -2018
Nguyễn Huy Tú, cho mk hỏi sao câu a bt đó lại bằng 0 vậy ? Mk ko hiểu lắm
\(\dfrac{72-x}{7}=\dfrac{x-4}{9}\)
\(\Rightarrow9\left(72-x\right)=7\left(x-4\right)\)
\(\Rightarrow648-9x=2x-28\)
\(\Rightarrow11x-28=648\)
\(\Rightarrow11x=676\Rightarrow x=\dfrac{676}{11}\)
\(\dfrac{37-x}{x+13}=\dfrac{3}{7}\)
\(\Rightarrow7\left(37-x\right)=3\left(x+13\right)\)
\(\Rightarrow259-7x=3x+39\)
\(\Rightarrow10x+39=259\)
\(\Rightarrow10x=220\Rightarrow x=22\)
\(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)
\(\Rightarrow\left(x+4\right)^2=100\)
\(\Rightarrow\left(x+4\right)^2=\pm10^2\)
\(\Rightarrow\left[{}\begin{matrix}x+4=10\Rightarrow x=6\\x+4=-10\Rightarrow x=-14\end{matrix}\right.\)
\(\dfrac{x-1}{x+2}=\dfrac{x-2}{x+3}\)
\(\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)\left(x+2\right)\)
\(\Rightarrow x\left(x+3\right)-1\left(x+3\right)=x\left(x+2\right)-2\left(x+2\right)\)
\(\Rightarrow x^2+3x-x-3=x^2+2x-2x-4\)
\(\Rightarrow x^2+2x-3=x^2-4\)
\(\Rightarrow2x-3=-4\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\dfrac{1}{2}\)
\(\left(2x+y^3\right)^2=4x^2+4xy^3+y^6\)
\(\left(\dfrac{1}{2}x-y\right)^2=\dfrac{1}{2}x^2-xy+y^2\)
\(\left(xy+5\right)^2=xy^2+10xy+25\)
\(\left(2y^2-3\right)^2=4y^4-12y^2+9\)
Các câu sau làm tương tự nha,dựa vào HĐT đó
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne2\\x\ne4\end{matrix}\right.\)
\(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\)
\(\Leftrightarrow\left(x-3\right).\left(x-4\right)+\left(x-2\right)^2=-\left(x-2\right).\left(x-4\right)\)
\(\Leftrightarrow3x^2-17x+24=0\)
\(\Leftrightarrow3x^2-9x-8x+24=0\)
\(\Leftrightarrow\left(3x-8\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-8=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=3\end{matrix}\right.\left(\text{thỏa}\right)\)
\(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\left(x\ne\left\{2;4\right\}\right)\\ =>\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=-1\\ =>x^2-3x-4x+12+x^2-4x+4=-\left(x-2\right)\left(x-4\right)\\ =>2x^2-11x+16=-x^2+6x-8\\ =>3x^2-17x+24=0\\ =>\left(x-3\right)\left(3x-8\right)=0\\ =>\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\left(TMDK\right)\)