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Đặt \(2n+2017=a^2;n+2019=b^2\)
\(\Rightarrow2n+4038=2b^2\)
\(\Rightarrow2b^2-a^2=2021\)
\(\Leftrightarrow\left(\sqrt{2b}-a\right)\left(\sqrt{2b}+a\right)=2021=1\cdot2021=47\cdot43\)
Tự xét nốt nha
\(\frac{1}{a}+\frac{1}{b}=\frac{1}{2019}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{1}{2019}\)
\(\Leftrightarrow2019a+2019b-ab=0\)
\(\Leftrightarrow ab-2019a-2019b=0\)
\(\sqrt{a+b}=\sqrt{a-2019}+\sqrt{b-2019}\)
\(\Leftrightarrow a+b=a-2019+b-2019+2\sqrt{\left(a-2019\right)\left(b-2019\right)}\)
\(\Leftrightarrow2\sqrt{ab-2019a-2019b+2019^2}=2\cdot2019\)
\(\Leftrightarrow2\cdot2019=2\cdot2019\) ( LUÔN OK THEO COOL KID ĐZ )
P/S:SORRY NHA.LÚC CHIỀU BẬN VÀI VIỆC NÊN KO ONL DC:(((
a) \(25^{n+1}-25^n=25^n\left(25-1\right)=25^n.4⋮25.4=100\)
b) \(n^2\left(n-1\right)-2n\left(n-1\right)=\left(n^2-2n\right)\left(n-1\right)\)
\(=n\left(n-1\right)\left(n-2\right)\)
Tích 3 số tự nhiên liên tiếp chia hết cho 6 nên \(n^2\left(n-1\right)-2n\left(n-1\right)⋮6\)
c) \(n^3-n=n\left(n^2-1\right)=\left(n-1\right)n\left(n+1\right)\)
Tích 3 số tự nhiên liên tiếp chia hết cho 6 nên \(n^3-n⋮6\)
\(\left(2n+3\right)^2-\left(2n-1\right)^2=4n^2+12n+9-4n^2+4n-1=16n+8=8\left(2n+1\right)⋮8\)
\(\left(2n+3\right)^2-\left(2n-1\right)^2\)
\(=\left(2n+3-2n+1\right)\left(2n+3+2n-1\right)\)
\(=4\left(4n-2\right)\)
\(=8\left(2x-1\right)\) Vì \(8⋮8\)
\(\Rightarrow8\left(2n-1\right)⋮(ĐPCM)\)
3/
a/ \(A=\left(x-y\right)^2+\left(x+y\right)^2.\)
\(A=\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)\)
\(A=x^2-2xy+y^2+x^2+2xy+y^2\)
\(A=2x^2+2y^2\)
b/ \(B=\left(2a+b\right)^2-\left(2a-b\right)^2\)
\(B=\left(4a^2+4ab+b^2\right)-\left(4a^2-4ab+b^2\right)\)
\(B=4a^2+4ab+b^2-4a^2+4ab-b^2\)
\(B=8ab\)
c/ \(C=\left(x+y\right)^2-\left(x-y\right)^2\)
\(C=\left(x^2+2xy+y^2\right)-\left(x^2-2xy+y^2\right)\)
\(C=x^2+2xy+y^2-x^2+2xy-y^2\)
\(C=4xy\)
d/ \(D=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(D=\left(4x^2-4x+1\right)-2\left(4x^2-12x+9\right)+4\)
\(D=4x^2-4x+1-8x^2+24x-18+4\)
\(D=-4x^2+20x-13\)
\(a,\left(x+2\right)^2=x^2+4x+4\)
\(b,\left(x-1\right)^2=x^2-2x+1\)
\(c,\left(x^2+y^2\right)^2=x^4+2x^2y^2+y^4\)
\(d,\left(x^3+2y^2\right)^2=x^6+4x^3y^2+4y^4\)
Bài 1:
a) Đặt \(6x+7=y\)
\(PT\Leftrightarrow y^2\left(y-1\right)\left(y+1\right)=72\)
\(\Leftrightarrow y^4-y^2-72=0\)
\(\Leftrightarrow\left(y^2-9\right)\left(y^2+8\right)=0\)
Mà \(y^2+8>0\left(\forall y\right)\)
\(\Rightarrow y^2-9=0\Leftrightarrow\left(y-3\right)\left(y+3\right)=0\Leftrightarrow\left(6x+4\right)\left(6x+10\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}6x+4=0\\6x+10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\x=-\frac{5}{3}\end{cases}}\)
b) đk: \(x\ne\left\{-4;-5;-6;-7\right\}\)
\(PT\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow x^2+11x+28=54\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-13\\x=2\end{cases}}\)
Bài 2 không tiện vẽ hình nên thôi nhờ godd khác:)
Bài 3:
Ta có:
\(a_n=1+2+3+...+n\)
\(a_{n+1}=1+2+3+...+n+\left(n+1\right)\)
\(\Rightarrow a_n+a_{n+1}=2\cdot\left(1+2+3+...+n\right)+\left(n+1\right)\)
\(=2\cdot\frac{n\left(n+1\right)}{2}+n+1\)
\(=n^2+n+n+1=\left(n+1\right)^2\)
Là SCP => đpcm
a) ( 3 - x )( x2 + 2x - 7 ) + ( x - 3 )( x2 + x - 5 )
= ( 3 - x )( x2 + 2x - 7 ) - ( 3 - x )( x2 + x - 5 )
= ( 3 - x )( x2 + 2x - 7 - x2 - x + 5 )
= ( 3 - x )( x - 2 )
b) ( x - 5 )2 + 3( 5 - x )
= ( x - 5 )2 - 3( x - 5 )
= ( x - 5 )( x - 5 - 3 ) = ( x - 5 )( x - 8 )
c) 2x( x - 1 )2 - ( 1 - x )3
= 2x( 1 - x )2 - ( 1 - x )3
= ( 1 - x )2( 2x - 1 + x ) = ( 1 - x )2( 3x - 1 )
d) x2 + 8x + 16 = ( x + 4 )2
e) x2 - 4xy + 4y2 = ( x - 2y )2
g) 4x2 - 25y2 = ( 2x )2 - ( 5y )2 = ( 2x - 5y )( 2x + 5y )
h) 25( x + 1 )2 - 4( x - 3 )2
= 52( x + 1 )2 - 22( x - 3 )2
= ( 5x + 5 )2 - ( 2x - 6 )2
= ( 5x + 5 - 2x + 6 )( 5x + 5 + 2x - 6 )
= ( 3x + 11 )( 7x - 1 )
i) x3 + 27 = ( x + 3 )( x2 - 3x + 9 )
k) 8x3 - 125 = ( 2x )3 - 53 = ( 2x - 5 )( 4x2 + 10x + 25 )
l) x3 + 6x2 + 12x + 8 = ( x + 2 )3
m) -x3 + 9x2 - 27x + 27 = -( x3 - 9x2 + 27x - 27 ) = -( x - 3 )3
2. Câu hỏi của Đình Hiếu - Toán lớp 7 - Học toán với OnlineMath