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\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=\dfrac{a+b+c+d}{a+b+c+d}=1\\ \Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=d\\d=a\end{matrix}\right.\Rightarrow a=b=c=d\\ \Rightarrow VT=\left(\dfrac{2019a+2020a-2021a}{2019a+2020a-2021a}\right)^3=1^3=1=\dfrac{a^2}{a\cdot a}=VP\)
1) Ta có : \(\frac{2016a+b+c+d}{a}=\frac{a+2016b+c+d}{b}=\frac{a+b+2016c+d}{c}=\frac{a+b+c+2016d}{d}\)
Trừ 4 vế với 2015 ta được : \(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
Nếu a + b + c + d = 0
=> a + b = -(c + d)
=> b + c = (-a + d)
=> c + d = -(a + b)
=> d + a = (-b + c)
Khi đó M = (-1) + (-1) + (-1) + (-1) = - 4
Nếu a + b + c + d\(\ne0\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=\frac{1}{d}\Rightarrow a=b=c=d\)
Khi đó M = 1 + 1 + 1 + 1 = 4
2) a) Ta có : \(\hept{\begin{cases}\left|x+2013\right|\ge0\forall x\\\left(3x-7\right)^{2004}\ge0\forall y\end{cases}\Rightarrow\left|x+2013\right|+\left(3x-7\right)^{2014}\ge0}\)
Dấu "=" xảy ra \(\hept{\begin{cases}x+2013=0\\3y-7=0\end{cases}\Rightarrow\hept{\begin{cases}x=-2013\\y=\frac{7}{3}\end{cases}}}\)
b) 72x + 72x + 3 = 344
=> 72x + 72x.73 = 344
=> 72x.(1 + 73) = 344
=> 72x = 1
=> 72x = 70
=> 2x = 0 => x = 0
c) Ta có :
\(\frac{7}{2x+2}=\frac{3}{2y-4}=\frac{5}{x+4}\Leftrightarrow\frac{7}{2x+2}=\frac{3}{2y-4}=\frac{10}{2x+8}=\frac{7-10}{2x+2-2x-8}=\frac{1}{2}\)(dãy tỉ số bằng nhau)
=> 2x + 2 = 14 => x = 6 ;
2y - 4 = 6 => y = 5 ;
6 + 5 + z = 17 => z = 6
Vậy x = 6 ; y = 5 ; z = 6
3) a) Ta có : \(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{a+b+c-a+b-c}{a+b-c-a+b+c}=\frac{2b}{2b}=1\)(dãy ti số bằng nhau)
=> a + b + c = a + b - c => a + b + c - a - b + c = 0 => 2c = 0 => c = 0;
Lại có : \(\frac{a+b+c}{a+b-c}-1=\frac{a-b+c}{a-b-c}-1\Leftrightarrow\frac{2c}{a+b-c}=\frac{2c}{a-b-c}\Rightarrow a+b-c=a-b-c\) => b = 0
Vậy c = 0 hoặc b = 0
c) Ta có : \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b+b+c+a+c}{c+a+b}=2\)(dãy tỉ số bằng nhau)
=> \(\hept{\begin{cases}a+b=2c\\b+c=2a\\a+c=2b\end{cases}}\)
Khi đó P = \(\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{b}{a}\right)=\frac{b+c}{b}.\frac{c+a}{c}=\frac{a+b}{a}=\frac{2a.2b.2c}{abc}=8\)
Vậy P = 8
2. b) \(7^{2x}+7^{2x+3}=344\)
\(7^{2x}\cdot\left(1+7^3\right)=344\)
\(7^{2x}\cdot\left(1+343\right)=344\)
\(7^{2x}\cdot344=344\)
\(7^{2x}=1\)
\(7^{2x}=7^0\)
\(2x=0\)
\(x=0\)
ADTCSTSBN , ta được :
\(\frac{3}{a+b}=\frac{2}{b+c}=\frac{1}{c+a}=\frac{6}{2\left(a+b+c\right)}=\frac{3}{a+b+c}\)
\(\Rightarrow a+b=a+b+c\)\(\Rightarrow c=0\)
\(P=\frac{a+b-2019.0}{a+b+2018.0}=\frac{a+b}{a+b}=1\)
Vậy P = 1
ta co
3/a+b=3/b+c=3/c+a
=>1:3/a+b=1:2/b+c=1:1/c+a
=>a+b/3=b+c/2=c+a/1
ap dung DTSBN, ta có
a+b/3=b+c/2=c+a/1+(a+b)+(b+c)+(c+a)/3+2+1=2a+2b+2c/6=2.(a+b+c)/6=a+b+c/3
vi a+b/3=a+b+c/3
=>a+b=a+b+c
=>c=0
=>p=a+b-2019.0/a+b+2018.0
=>p=a+b/a+b
=>p=1
KL
Với \(a+b+c+d=0\)
\(\Rightarrow a+b=-\left(c+d\right);b+c=-\left(d+a\right);c+d=-\left(a+b\right);d+a=-\left(b+c\right)\)
Khi đó \(M=-1-1-1-1=-4\)
Với \(a+b+c+d\ne0\)
Áp dụng dãy tỉ số bằng nhau
\(\frac{2019a+b+c+d}{a}=\frac{a+2019b+c+d}{b}=\frac{a+b+2019c+d}{c}=\frac{a+b+c+2019d}{d}\)
\(=\frac{2022\left(a+b+c+d\right)}{a+b+c+d}=2022\)
\(\Rightarrow a=b=c=d\)
\(\Rightarrow M=4\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\Rightarrow\frac{2019a^2}{2019c^2}=\frac{2020b^2}{2020d^2}=\)
\(=\frac{2019a^2+2020b^2}{2019c^2+2020d^2}=\frac{2019a^2-2020b^2}{2019c^2-2020d^2}\Rightarrow\frac{2019a^2+2020b^2}{2019a^2-2020b^2}=\frac{2019c^2+2020d^2}{2019c^2-2020d^2}\)
Bạn ơi tham khảo thử cách này nhé !
Từ \(\frac{a}{b}=\frac{c}{d}\)( bài cho )
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó :
+) \(\frac{2019a^2+2020b^2}{2019a^2-2020b^2}=\frac{2019\left(bk\right)^2+2020b^2}{2019\left(bk\right)^2-2020b^2}=\frac{b^2\left(2019k^2+2020\right)}{b^2\left(2019k^2-2020\right)}=\frac{2019k^2+2020}{2019k^2-2020}\)
+) \(\frac{2019c^2+2020d^2}{2019c^2-2020d^2}=\frac{2019\left(dk\right)^2+2020d^2}{2019\left(dk\right)^2-2020d^2}=\frac{d^2\left(2019k^2+2020\right)}{d^2\left(2019k^2-2020\right)}=\frac{2019k^2+2020}{2019k^2-2020}\)
ta có\(\left|x+y-5\right|\ge0\)
\(\left(y-2\right)^8\ge0\)
để biểu thức = 0 thì 2 biểu thức trên =0
\(tacó\)\(x+y=5\)
\(y-2=0\Rightarrow y=2\)
\(x+2=5\Rightarrow x=3\)
x=2;y=3
