Em kiểm tra lại đề nhé!

Nếu đề đúng:

\(10\cdot\frac{4^6\cdot9^5+69\cdot120}{8^4\cdot3^{12}-6^{11}}\)

\(=10\cdot\frac{\left(2^2\right)^6\cdot\left(3^2\right)^5+\left(3.23\right)\cdot\left(2.2.2.3.5\right)}{\left(2^3\right)^4\cdot3^{12}-\left(2.3\right)^{11}}\)

\(=10\cdot\frac{2^{12}\cdot3^{10}+3^2.2^3.23.5}{2^{12}\cdot3^{12}-2^{11}.3^{11}}\)

\(=2.5\cdot\frac{2^3\cdot3^2\left(2^9.3^8+23.5\right)}{2^{11}.3^{11}\left(2.3-1\right)}\)

\(=\frac{2^4\cdot3^2.5\left(2^9.3^8+23.5\right)}{2^{11}.3^{11}.5}=\frac{3359347}{2^7.3^9}\)

\(B=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n.\left(n+2\right)}\right)\)

\(=\left(\frac{1.3+1}{1.3}\right).\left(\frac{2.4+1}{2.4}\right).\left(\frac{3.5+1}{3.5}\right)...\left(\frac{n.\left(n+2\right)+1}{n.\left(n+2\right)}\right)\)

\(=\left(\frac{2^2}{1.3}\right).\left(\frac{3^2}{2.4}\right).\left(\frac{4^2}{3.5}\right)...\left(\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\right)\)

\(=\frac{2.3.4...\left(n+1\right)}{1.2.3...n}.\frac{2.3.4...\left(n+1\right)}{3.4.5...\left(n+2\right)}\)

\(=\frac{\left(n+1\right)}{1}.\frac{2}{\left(n+2\right)}\)

\(=\frac{2.\left(n+1\right)}{1.\left(n+2\right)}=2.\frac{n+1}{n+2}< 2\)(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