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hỏi mỗi từng câu 1 thôi nhé ! Vậy mình giải cho . Mình k có ý kiếm GP + SP đâu . Nhưng nhìn 8 câu này hoa hết cả mắt :v
Đúng thật. Tớ nhìn cũng thấy ngán mà. Nhiều quá nên hơi nản
a) Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\\\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\end{matrix}\right.\Rightarrowđpcm\)
b) \(\dfrac{x-1}{2017}+\dfrac{x-2}{2016}=\dfrac{x-3}{2015}+\dfrac{x-4}{2014}\)
\(\Rightarrow\left(\dfrac{x-1}{2017}-1\right)+\left(\dfrac{x-2}{2016}-1\right)=\left(\dfrac{x-3}{2015}-1\right)+\left(\dfrac{x-4}{2014}-1\right)\)\(\Rightarrow\dfrac{x-2018}{2017}+\dfrac{x-2018}{2016}=\dfrac{x-2018}{2015}+\dfrac{x-2018}{2014}\)
\(\Rightarrow\left(x-2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
vì \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\) nên \(x-2018=0\Leftrightarrow x=2018\)
Từ \(\dfrac{x}{y}=\dfrac{9}{7}\)ta có : \(x=\dfrac{9y}{7}\)(1) ;
Từ \(\dfrac{y}{z}=\dfrac{7}{3}\)ta có: \(z=\dfrac{3y}{7}\)(2);
Thay (1) và (2) vào biểu thức trên ta có:
\(\left(\dfrac{9y}{7}\right)^2-\left(\dfrac{9y^2}{7}\right)+\left(\dfrac{3y}{7}\right)^2=27=>\dfrac{81y^2}{49}-\dfrac{63y^2}{49}+\dfrac{9y^2}{49}=27\)
\(=>\dfrac{27y^2}{49}=27=>27y^2=27.49=1323\)
\(=>y^2=1323:27=49=>y=7;-7\)
Lần lượt thay y =7; -7 vào hệ thức ta tìm được:
\(y=7;x=9;z=3\)và \(y=-7;x=-9;z=-3\)
CHÚC BẠN HỌC TỐT...
Đặt \(\dfrac{a+b}{3}=\dfrac{b+c}{4}=\dfrac{c+a}{5}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=3k\\b+c=4k\\c+a=5k\end{matrix}\right.\)
\(\Rightarrow2\left(a+b+c\right)=12k\)
\(\Rightarrow a+b+c=6k\) \(\Rightarrow\left\{{}\begin{matrix}a=2k\\b=k\\c=3k\end{matrix}\right.\)
Thay a = 2k , b = k , c= = 3k vào biểu thức M , ta có :
M = 10.2k + k - 7.3k + 2017 = 21k - 21k + 2017 = 2017
Câu 7:
x=2014 nên x-1=2013
\(A=x^{2014}-x^{2013}\left(x-1\right)-x^{2012}\left(x-1\right)-...-x\left(x-1\right)+1\)
\(=x^{2014}-x^{2014}+x^{2013}-x^{2013}+x^{2012}-...-x^2+x+1\)
=x+1
=2014+1=2015
Nhiều quá, từng bài 1 nhé, bài nào làm được, tớ sẽ cố gắng.
bài 2:
a) \(x>2x\Leftrightarrow x-2x>0\Leftrightarrow-x>0\Leftrightarrow x< 0\)
Kl: x<0
b) \(a+x< a\Leftrightarrow x< 0\)
Kl: x<0
c) \(x^3>x^2\Leftrightarrow x^3-x^2>0\Leftrightarrow x^2\left(x-1\right)>0\) (*)
Mà x^2 > 0 \(\Rightarrow\) (*) \(\Leftrightarrow x-1>0\Leftrightarrow x>1\)
Kl: x>1
Câu 4:
a) \(1-2x< 7\Leftrightarrow2x>-6\Leftrightarrow x>3\)
Kl: x>3
b) \(\left(x-1\right)\left(x-2\right)>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\\x-2< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x< 1\\x< 2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>2\\x< 1\end{matrix}\right.\)
Kl: x>2 hoặc x<1
c) \(\left(x-2\right)^2\left(x+1\right)\left(x+4\right)< 0\Leftrightarrow\left(x+1\right)\left(x+4\right)< 0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\\x+4< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1< 0\\x+4>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-1\\x< -4\end{matrix}\right.\\\left\{{}\begin{matrix}x< -1\\x>-4\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-1< x< -4\left(vô-lý\right)\\-4< x< -1\end{matrix}\right.\) \(\Leftrightarrow-4< x< -1\)
Kl: -4<x<-1
d) ĐK: x khác 9\(\dfrac{x^2\left(x+3\right)}{x-9}< 0\Leftrightarrow x^2\left(x+3\right)\left(x-9\right)< 0\Leftrightarrow\left(x+3\right)\left(x-9\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3>0\\x-9< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+3< 0\\x-9>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-3\\x< 9\end{matrix}\right.\\\left\{{}\begin{matrix}x< -3\\x>9\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-3< x< 9\left(N\right)\\9< x< -3\left(vô-lý\right)\end{matrix}\right.\) \(\Leftrightarrow-3< x< 9\)
Kl: -3<x<9
e) Đk: x khác 0
\(\dfrac{5}{x}< 1\Leftrightarrow\dfrac{5}{x}< \dfrac{5}{5}\Leftrightarrow x>5\left(N\right)\)
KL: x >5
f) ĐK: x khác 1
\(\dfrac{2x-5}{x-1}< 0\Leftrightarrow\left(2x-5\right)\left(x-1\right)< 0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5>0\\x-1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5< 0\\x-1>0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{5}{2}\\x< 1\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x>1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{2}< x< 1\left(vô-lý\right)\\1< x< \dfrac{5}{2}\left(N\right)\end{matrix}\right.\)
Kl: 1< x< 5/2
a) Ta có : 7y=4z
=> \(\dfrac{y}{z}=\dfrac{4}{7}\)
Mà \(\dfrac{x+y}{t+z}=\dfrac{4}{7}\) nên \(\dfrac{x+y}{t+z}=\dfrac{y}{z}=\dfrac{4}{7}\)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{x+y}{t+z}=\dfrac{y}{z}=\dfrac{4}{7}=\dfrac{x+y-y}{t+z-z}=\dfrac{x}{t}\)
Vậy \(\dfrac{x}{t}=\dfrac{4}{7}\)