Cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)và \(a+b+c\ne0\).CMR: \(\left(19a+5b+1890\right)^{2019}=1914^{2019}.a^{2018}.b\)
cho a/b=b/c=c/d và a+b+c khác 0 cmr (19a+5b+1890)^2019=1914^2019.a^2018.b
có\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)CMR\(\frac{\left(19a+5b+1980c\right)^{2003}}{1914^{2003}.a^{2001}.b^2}\)
a) Cho các số dương a,b,c,d; c khác d và \(\frac{a}{b}\)=\(\frac{c}{d}\). Chứng minh rằng : \(\frac{\left(a^{2018}+b^{2018}\right)^{2019}}{\left(c^{2018}+d^{2018}\right)^{2019}}\)=\(\frac{\left(a^{2019}-b^{2019}\right)^{2018}}{\left(c^{2019}-d^{2019}\right)^{2018}}\)
b) Cho biết |3x + 2y| + |5z - 7x| + \(\left(xy+yz+xz-500\right)^{2022}\)= 0 . Tính giá trị : \(A=\left(3x-y-z\right)^{2021}\)
Các bạn giải giúp mik nhé. Mik cần gấp lắm. Ai giải trc mik sẽ tick cho
Cho a, b, c khác 0 và \((a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=1\)
Tính \(P=\left(a^{2018}-b^{2018}\right)\left(b^{2019}+c^{2019}\right)\left(c^{2019}-a^{2019}\right)\).
~help me~
cho \(\frac{a+b}{2018}=\frac{b+c}{2019}=\frac{c+a}{2020}\)
CMR \(\left(b-c\right)^2=4\left(b-a\right)\left(a-c\right)\)
Ta có :
\(\frac{a+b-b-c}{2018-2019}=\frac{a-c}{-1}\)
\(\frac{b+c-c-a}{2019-2020}=\frac{b-a}{-1}\)
\(\frac{b-c}{2018-2020}=\frac{b-c}{-2}\)
Đặt \(\frac{a-c}{-1}=\frac{b-a}{-1}=\frac{b-c}{-2}=k\left(k\ne0\right)\)
\(\Rightarrow\hept{\begin{cases}\frac{a-c}{-1}=k\\\frac{b-a}{-1}=k\\\frac{b-c}{-2}=k\end{cases}\Rightarrow\hept{\begin{cases}a-c=-k\\b-a=-k\\b-c=k.\left(-2\right)\end{cases}}}\)
\(\Rightarrowđpcm\)
Cho a, b, c \(\ne\) và \((a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=1\)
Tính giá trị biểu thức: \(P=\left(a^{2018}-b^{2018}\right)\left(b^{2019}+c^{2019}\right)\left(c^{2020}-d^{2020}\right)\).
\(Cho\) \(\frac{a}{2018}=\frac{b}{2019}=\frac{c}{2020}\)CMR
\(4\left(a-b\right)\left(b-c\right)=\left(a-c\right)^2\)
Đặt \(\frac{a}{2018}=\frac{b}{2019}=\frac{c}{2020}=k\)
\(\Rightarrow a=2018k\), \(b=2019k\), \(c=2020k\)
Ta có: \(4\left(a-b\right)\left(b-c\right)=4\left(2018k-2019k\right)\left(2019k-2020k\right)\)
\(=4.\left(-k\right).\left(-k\right)=4k^2=\left(2k\right)^2\)
Ta lại có: \(\left(a-c\right)^2=\left(2018k-2020k\right)^2=\left(-2k\right)^2=\left(2k\right)^2\)
Vậy \(4\left(a-b\right)\left(b-c\right)=\left(a-c\right)^2\)
Đặt \(\frac{a}{2018}=\frac{b}{2019}=\frac{c}{2020}=k\Rightarrow\hept{\begin{cases}a=2018k\\b=2019k\\c=2020k\end{cases}}\)
Thế vị trí tương ứng ta được :
VT = 4( a - b )( b - c )
= 4( 2018k - 2019k )( 2019k - 2020k )
= 4(-k)(-k)
= 4k2
VP = ( a - c )2
= ( 2018k - 2020k )2
= ( -2k )2
= 4k2
=> VT = VP
=> đpcm
cho:\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\) \(\left(a+b+c\ne0\right)\)
Biết a = 2019,tìm b và c
a=2019 =>do a/b=c/a => bc=a2=20192=>b2.c2=20194
doa/b=b/c => b2 =ac => b2=2019c => b2c2=2019c3
=> c3=20193 => c= 2019 => b=2019
Áp dụng tính chất dãy ti số = nhau, ta có :
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)
Khi đó : \(\frac{a}{b}=1\Rightarrow a=b\)mà \(a=2019\Rightarrow b=2019\)
\(\frac{c}{a}=1\Rightarrow c=a\) mà \(a=2019\Rightarrow c=2019\)
Vậy b = 2019 và c = 2019
Cho \(a,b,c\ne0\)có \(a+b=c+\frac{1}{2018}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+2018\)
Chứng minh \(P=a^{2019}+b^{2019}+c^{2019}\)