\(10\cdot\frac{4^6\cdot9^5+69\cdot120}{8^4\cdot3^{12}-6^{11}}\) HEPL ME !!!!!!!!!!!!!!!!!!!!
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\(B=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n.\left(n+2\right)}\right)\)
\(=\left(\frac{1.3+1}{1.3}\right).\left(\frac{2.4+1}{2.4}\right).\left(\frac{3.5+1}{3.5}\right)...\left(\frac{n.\left(n+2\right)+1}{n.\left(n+2\right)}\right)\)
\(=\left(\frac{2^2}{1.3}\right).\left(\frac{3^2}{2.4}\right).\left(\frac{4^2}{3.5}\right)...\left(\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\right)\)
\(=\frac{2.3.4...\left(n+1\right)}{1.2.3...n}.\frac{2.3.4...\left(n+1\right)}{3.4.5...\left(n+2\right)}\)
\(=\frac{\left(n+1\right)}{1}.\frac{2}{\left(n+2\right)}\)
\(=\frac{2.\left(n+1\right)}{1.\left(n+2\right)}=2.\frac{n+1}{n+2}< 2\)(vì \(\frac{n+1}{n+2}< 1\))
Vậy B < 2
Ta có:
\(1+\frac{1}{1.3}=\frac{4}{1.3}=\frac{2^2}{1.3}\)
\(1+\frac{1}{2.4}=\frac{9}{2.4}=\frac{3^2}{2.4}\)
\(1+\frac{1}{3.5}=\frac{16}{3.5}=\frac{4^2}{3.5}\)
...
\(1+\frac{1}{n\left(n+2\right)}=\frac{n^2+2n+1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
=>
\(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{\left(n+1\right)^2}{n\left(n+2\right)}=\frac{2^2.3^2.4^2...\left(n+1\right)^2}{1.2.3^2.4^2...\left(n+1\right)\left(n+2\right)}=\frac{2.\left(n+1\right)}{1.\left(n+2\right)}\)
\(=\frac{2\left(n+2\right)-2}{n+2}=2-\frac{2}{n+2}< 2\)
Vậy B < 2
| 4x - 1 | + 4x =5
TH1: \(4x-1\ge0\)
\(4x-1+4x=5\)
\(4x+4x=5+1\)
\(8x=6\)
\(x=\frac{3}{4}\)( thỏa mãn \(4x-1\ge0\))
TH2: 4x - 1 <0
\(-\left(4x-1\right)+4x=5\)
\(-4x+1+4x=5\)
\(1=5\) ( vô lí)
Vậy \(x=\frac{3}{4}\)
1. A = 100
2. B = 2098
mik ko biết có đúng ko đâu nhé vì mình nhowfbanj làm cho rồi viết vô đây mà ahihi
\(2x^2+x-3=2\sqrt{2x+3}\left(ĐKXĐ:x\ge-\frac{3}{2}\right)\)
Đặt \(\sqrt{2x+3}=2y-1\left(y\ge\frac{1}{2}\right)\)
Ta có hpt \(\hept{\begin{cases}2x^2+x-3=2\left(2y-1\right)\\\left(2y-1\right)^2=2x+3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x^2+x-3-4y+2=0\\2y^2-2y-x-1=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x^2+x-4y=1\left(1\right)\\2y^2-2y-x=1\left(2\right)\end{cases}}\)
Lấy (1)-(2)\(\Leftrightarrow2\left(x^2-y^2\right)+x-4y+2y+x=0\)
\(\Leftrightarrow2\left(x-y\right)\left(x+y\right)+2\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y+1\right)=0\)
Nhận thấy \(x+y+1\ge-\frac{3}{2}+\frac{1}{2}+1=0\)
Dấu "=" xảy ra <=>x=-3/2;y=1/2(thỏa mãn)
Nếu x=y
\(\Rightarrow\sqrt{2x+3}=2y-1\Leftrightarrow\sqrt{2x+3}=2x-1\)
\(\Leftrightarrow\hept{\begin{cases}2x+3=4x^2-4x+1\left(3\right)\\x\ge\frac{1}{2}\end{cases}}\)
\(\left(3\right)\Leftrightarrow4x^2-6x-2=0\Leftrightarrow2x^2-3x-1=0\)
Dùng đenta giải ra nghiệm rồi kết luận nha
a, 3.(7+x)=88-64
3.(7+x)=24
7+x=24:3
7+x=8
x=1
b,7x=114-44
7x=70
x=70:7
x=10
c,31x-41=114
31x=114+41
31x=155
x=5
k như đã nói nha 3 cái nha
3.(7+x)=88-64 7x=114-44. 31x-41=57.2
3.(7+x)=24 7x=70 31x-41=114
7+x=24:3 x=70:7 31x =114+41
7+x=8 x=10 31x =155
x=8-7 x =155:31
x = 5
x=1
\(\Leftrightarrow\left(1-3x\right)^3=\left(-5\right)^3\)
\(\Leftrightarrow1-3x=-5\)
\(\Leftrightarrow3x=1-\left(-5\right)\)
\(\Leftrightarrow3x=6\)
\(\Rightarrow x=2\)
a, ta có (-5)3=-125
=>1-3x=-5
3x=1+5
3x=6
x=2
b,34-x=27
81-x=27
x=54
nha
a) \(2^{x-1}.3=48\)
\(2^{x-1}=16\)
\(2^{x-1}=2^4\)
\(x-1=4\)
\(x=5\)
b.
\(\frac{\left(-3\right)^x}{81}=-27\)
\(\frac{\left(-3\right)^x}{3^4}=\left(-3\right)^3\)
\(\frac{\left(-3\right)^x}{\left(-3\right)^4}=\left(-3\right)^3\)
\(\left(-3\right)^x=\left(-3\right)^4.\left(-3\right)^3\)
\(\left(-3\right)^x=\left(-3\right)^7\)
x=7
Em kiểm tra lại đề nhé!
Nếu đề đúng:
\(10\cdot\frac{4^6\cdot9^5+69\cdot120}{8^4\cdot3^{12}-6^{11}}\)
\(=10\cdot\frac{\left(2^2\right)^6\cdot\left(3^2\right)^5+\left(3.23\right)\cdot\left(2.2.2.3.5\right)}{\left(2^3\right)^4\cdot3^{12}-\left(2.3\right)^{11}}\)
\(=10\cdot\frac{2^{12}\cdot3^{10}+3^2.2^3.23.5}{2^{12}\cdot3^{12}-2^{11}.3^{11}}\)
\(=2.5\cdot\frac{2^3\cdot3^2\left(2^9.3^8+23.5\right)}{2^{11}.3^{11}\left(2.3-1\right)}\)
\(=\frac{2^4\cdot3^2.5\left(2^9.3^8+23.5\right)}{2^{11}.3^{11}.5}=\frac{3359347}{2^7.3^9}\)