TÍNH NHANH
5^x+1=5^5-5^2
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Bài 1 :
Gọi \(A=5+5^2+5^3+...+5^{98}+5^{99}\\ 5A=5^2+5^3+5^4+...+5^{99}+5^{100}\\ 5A-A=\left(5^2+5^3+5^4+...+5^{99}+5^{100}\right)-\left(5+5^2+5^3+...+5^{98}+5^{99}\right)\\ 4A=5^{100}-5\\ A=\dfrac{5^{100}-5}{4}\)
Bài 2:
\(\left(12x-4\right)\cdot8^{2022}=4\cdot8^{2023}\\ 12x-4=4\cdot8^{2023}:8^{2022}\\ 12x-4=4\cdot8\\ 12x-4=32\\ 12x=36\\ x=3\)
5/6 x 1/7 - 1/7 x 1/6
= 1/7 x (5/6 - 1/6)
= 1/7 x 2/3
= 2/21
8/15 : 4/5 x 3/2
= 2/3 x 3/2
= 1
\(=\left(1,1+1,9\right)+\left(1,2+1,8\right)+\left(1,3+1,7\right)+\left(1,4+1,6\right)+1,5\)
\(=3+3+3+3+1,5\)
\(=13,5\)
Bài 2:
a: 2/6x5/3=10/18=5/9
b: 11/9x5/10=55/90=11/18
c: 3/9x6/8=1/3x3/4=1/4
d: 4/9x12/16=48/144=1/3
e: 25/15x6/7=5/3x6/7=30/21=10/7
f: 6/10x15/20=90/200=9/20
Bài 1
4/5 x 6/7= 24/35
2/9 x 1/2= 2/18= 1/9
1/2 x 8/3= 8/6= 4/3
7/9 x 6/5= 42/45= 14/15
8/7 x 5/9= 40/63
10/11 x 22/15= 220/165= 4/3
Bài 2
2/6 x 5/3= 1/3 x 5/3=5/9
11/9 x 5/10= 11/9 x 1/2= 11/18
3/9 x 6/8= 1/3 x 3/4 =3/12= 1/4
4/9 x 12/16= 4/9 x 3/4= 12/36= 1/3
25/15 x 6/7= 5/3 x 6/7= 30/21= 10/7
6/10 x 15/20= 3/5 x 3/4= 9/20
(2/3 x 4/5) x 5/6
= 8/15 x 5/6
= 4/9
(2/3 x 4/5) x 5/6
= (5/6 x 4/5) x 2/3
= 2/3 x 2/3
= 4/9
(1/2 + 1/3) x 1/5
= 5/6 x 1/5
= 1/6
(1/2 + 1/3) x 1/5
= (1/2 x 1/5) + (1/3 x 1/5)
= 1/10 + 1/15
= 1/6
4A6?
a: \(A=\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
\(=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\left(\dfrac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)\)
\(=\left(1-\sqrt{5}\right)\left(-1-\sqrt{5}\right)\)
\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=5-1=4\)
b: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)
\(B=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)
\(=\dfrac{1}{2\left(\sqrt{x}-1\right)}-\dfrac{1}{2\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=-\dfrac{2}{\sqrt{x}+1}\)
c: Khi x=9 thì \(B=\dfrac{-2}{\sqrt{9}+1}=\dfrac{-2}{3+1}=-\dfrac{2}{4}=-\dfrac{1}{2}\)
d: |B|=A
=>\(\left|-\dfrac{2}{\sqrt{x}+1}\right|=4\)
=>\(\dfrac{2}{\sqrt{x}+1}=4\) hoặc \(\dfrac{2}{\sqrt{x}+1}=-4\)
=>\(\sqrt{x}+1=\dfrac{1}{2}\) hoặc \(\sqrt{x}+1=-\dfrac{1}{2}\)
=>\(\sqrt{x}=-\dfrac{1}{2}\)(loại) hoặc \(\sqrt{x}=-\dfrac{3}{2}\)(loại)