A= (x+1)(x+2)(x+3)(x+4)-32
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có rút gọn phân số 35/6 về phân số đc ko ? Nếu có thì mn rút gọn cho mik nhé !
Ai xong tr tui k cho

Có: \(A=4\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=...........................\)
\(=\frac{3^{32}-1}{2}\)
\(B=3^{32-1}\)
=> \(A< B\)

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

a) \(x-2=-6\)
\(x=-6+2\)
\(x=-4\)
b) \(15-\left(x-7\right)=-21\)
\(x-7=36\)
\(x=43\)
c) \(4.\left(3x-4\right)-2=18\)
\(4\left(3x-4\right)=20\)
\(3x-4=5\)
\(3x=9\)
\(x=3\)
d) \(\left(3x-6\right)+3=32\)
\(3x-6=29\)
\(3x=29+6\)
\(3x=35\)
\(x=\frac{35}{3}\)
e) \(\left(3x-6\right).3=32\)
\(3x-6=\frac{32}{3}\)
\(3x=\frac{32}{3}+6\)
\(3x=\frac{50}{3}\)
\(x=\frac{50}{9}\)
f) \(\left(3x-6\right):3=32\)
\(3x-6=96\)
\(3x=102\)
\(x=34\)
g) \(\left(3x-6\right)-3=32\)
\(3x-6=35\)
\(3x=41\)
\(x=\frac{41}{3}\)
h) \(\left(3x-2^4\right).7^3=2.7^4\)
\(\left(3x-2^4\right)=2.7=14\)
\(\left(3x-16\right)=14\)
\(3x=14+16=30\)
\(x=10\)
i) \(\left|x\right|=\left|-7\right|\)
\(\left|x\right|=7\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)
k) \(\left|x+1\right|=2\)
\(\Rightarrow\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)
l) \(\left|x-2\right|=3\)
\(\Rightarrow\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)
m) \(x+\left|-2\right|=0\)
\(x+2=0\)
\(x=-2\)
o) \(72-3\left|x+1\right|=9\)
\(3\left|x-1\right|=63\)
\(\left|x-1\right|=21\)
\(\Rightarrow\orbr{\begin{cases}x-1=21\\x-1=-21\end{cases}\Rightarrow\orbr{\begin{cases}x=22\\x=-20\end{cases}}}\)
p) Ta có: \(\left|x-1\right|=3\)
\(\Rightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}}\)
mà \(x+1< 0\)
\(\Rightarrow x-1=-3\)
\(\Rightarrow x=-2\)
q) \(\left(x-2\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}}\)
hok tốt!!

a) (x + 1)(2x – 2) – 3 > –5x – (2x + 1)(3 – x)
⇔ 2x2 – 2x + 2x – 2 – 3 > –5x – (6x – 2x2 + 3 – x)
⇔ 2x2 – 5 ≥ –5x – 6x + 2x2 – 3 + x
⇔ 10x ≥ 2 ⇔ x ≥ 1/5
Tập nghiệm: S = {x | x ≥ 1/5}
b) (x – 3)2 + 4(2 – x) > x(x + 7)
⇔ x2 – 6x + 9 + 8 – 4x > x2 + 7x
⇔ –17x > –17
⇔ x < -17/-17
⇔ x < 1
Tập nghiệm: S = {x | x < 1}.

Bài 10:
a: Để A là phân số thì n+2<>0
hay n<>-2
b: Khi n=0 thì A=3/2
Khi n=2 thì A=3/(2+2)=3/4
Khi n=-7 thì A=3/(-7+2)=-3/5
Bài 9:
1)9/x = -35/105 2) 12/5 = 32/x 3)x/2 = 32/x x = 9. (-35)/105 x.12/5 = x.32/x 2x.x/2 = 2x.32/x
x = -3 x.12/5=32 xx = 2.32
x= 32:12/5 x^2 = 2.32
x = 40/3 x^2 = 64
x = 8
4) x-2/4 = x-1/5
5(x-2) = 4(x-1)
5x - 10 = 4x - 4
5x - 4x = 10 - 4
x = 6
Bài 10:Cho biểu thức A=3/n+2
a) Để A là phân số thì mẫu số phải khác 0
Do đó: n + 2 ≉ 0. Suy ra: n ≉ -2
b) Khi n = 0 thì A = 3/0+2 = 3/2
Khi n = 2 thì A = 3/2+2 = 3/4
Khi n = -7 thì A = 3/-7+2 = 3/-5

6:=(3/2)*(3/2)^2*(3/2)^4=(3/2)^7
7: =(1/2)^7*2^3*2^5*2^8=2^9
8: =(-1/7)^4*5^4=(-5/7)^4
9: =2^2*2^5:(2^3/2^4)
=2^7/2=2^6
10: =(1/7)^3*7^2=1/7

a) ĐKXĐ: \(x\ge0\)
Ta có: \(\left(x+3\sqrt{x}+2\right)\left(x+9\sqrt{x}+18\right)=168x\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+6\right)=168x\)
\(\Leftrightarrow\left(x+6\right)^2+12\sqrt{x}\left(x+6\right)-133=0\)
\(\Leftrightarrow\left(x+6\right)^2+19\sqrt{x}\left(x+6\right)-7\sqrt{x}\left(x+6\right)-133=0\)
\(\Leftrightarrow\left(x+6\right)\left(x+19\sqrt{x}+6\right)-7\sqrt{x}\left(x+19\sqrt{x}+6\right)=0\)
\(\Leftrightarrow\left(x-7\sqrt{x}+6\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=36\end{matrix}\right.\)
Dòng thứ 2 qua dòng thứ 3 anh làm chậm lại được không ạ, tại tắt quá e không hiểu
Ta có: \(A=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-32\)
\(\Leftrightarrow A=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-32\)
\(\Leftrightarrow A=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-32\)
Đặt \(a=x^2+5x+4\) \(\Rightarrow\)\(a+2=x^2+5x+6\)
Ta lại có: \(A=a.\left(a+2\right)-32\)
\(\Leftrightarrow A=a^2+2a-32\)
\(\Leftrightarrow A=a^2+2a-32\)
\(\Leftrightarrow A=\left(a^2+2a+1\right)-33\)
\(\Leftrightarrow A=\left(a+1\right)^2-\left(\sqrt{33}\right)^2\)
\(\Leftrightarrow A=\left(a+1-\sqrt{33}\right)\left(a+1+\sqrt{33}\right)\)
\(\Leftrightarrow A=\left(x^2+5x+5-\sqrt{33}\right)\left(x^2+5x+5+\sqrt{33}\right)\)
Ta có A = (x + 1)(x + 2)(x + 3)(x + 4) - 32
= [(x + 1)(x + 4)][(x + 2)(x + 3)] - 32
= (x2 + 5x + 4)(x2 + 5x + 6) - 32
= (x2 + 5x + 5 - 1)(x2 + 5x + 5 + 1) - 32
= (x2 + 5x + 5)2 - 1 - 32
= (x2 + 5x + 5)2 - 33 \(\ge-33\)
Dấu "=" xảy ra <=> x2 + 5x + 5 = 0
=> (x2 + 5x + 25/4) = 5/4
=> (x + 5/2)2 = 5/4
=> \(\orbr{\begin{cases}x+\frac{5}{2}=\sqrt{\frac{5}{4}}\\x+\frac{5}{2}=-\sqrt{\frac{5}{4}}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\sqrt{\frac{5}{4}}-\frac{5}{2}\\x=-\sqrt{\frac{5}{4}}-\frac{5}{2}\end{cases}}\)
Vậy Min A = -33 <=> \(x\in\left\{\sqrt{\frac{5}{4}}-\frac{5}{2};-\sqrt{\frac{5}{4}}-\frac{5}{2}\right\}\)