uỵttyutyutu

2) 11¹³ - 11¹² - 11¹¹

= 11¹¹⁺² - 11¹¹⁺¹ - 11¹¹

= 11¹¹.11² - 11¹¹.11 - 11¹¹

= 11¹¹(11² - 11 - 1)

= 11¹¹.109 \(⋮\) 109

vậy.........

mik ko biết nhưng hình như câu 1 sai đề bài hay sao ý

Đề ở mẫu thứ 3 là \(\sqrt{xz}+12\sqrt{z}+12\) mới đúng

Ta có: \(\sqrt{xyz}=12\)

\(\Rightarrow P=\dfrac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+\sqrt{xyz}}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{xyz}.\sqrt{z}}{\sqrt{xz}+\sqrt{xyz}.\sqrt{x}+\sqrt{xyz}}\)\(=\dfrac{1}{\sqrt{y}+1+\sqrt{yz}}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{yz}}{1+\sqrt{y}+\sqrt{yz}}=1\)

đậu xanh sửa thành z mak vẫn ghi x

a: \(=\dfrac{-2}{7}\cdot\dfrac{3}{2}=\dfrac{-3}{7}\)

b: \(=3\cdot\dfrac{7}{12}=\dfrac{7}{4}\)

c: \(=\dfrac{11}{12}\cdot\dfrac{16}{33}\cdot\dfrac{3}{5}=\dfrac{1}{3}\cdot\dfrac{4}{3}\cdot\dfrac{3}{5}=\dfrac{1}{3}\cdot\dfrac{4}{5}=\dfrac{4}{15}\)

1) \(12+\left(4-x\right)=-5\)

\(\Leftrightarrow12+4-x=-5\)

\(\Leftrightarrow-x=-5-12-4=-21\)

\(\Leftrightarrow x=21\)

2) \(\left|x-6\right|=5\)

\(\Leftrightarrow\left[\begin{matrix}x-6=5\\x-6=-5\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=11\\x=1\end{matrix}\right.\)

3) \(\left|x-3\right|=4\)

\(\Leftrightarrow\left[\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

4) \(x=-11\)

a, 12+(4-x)=-5
4-x=7
=>x=3
b,|x-6|=5
=>x-6=\(\pm6\)

Xét 2 TH:

\(\Rightarrow\left[\begin{matrix}x-6=6\\x-6=-6\end{matrix}\right. \)

\(\Rightarrow\left[\begin{matrix}x=12\\x=0\end{matrix}\right.\)

Vậy \(x\in\left\{0;12\right\}\)

c, |x-3|=4

x-3=±4

Xét 2 TH:

\(\Rightarrow\left[\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

Vậy \(x\in\left\{-1;7\right\}\)

d, 12+11+10+...+x=12

\(\Rightarrow11+10+...+x=0\)

Gọi n là số số hạng của vế trái.

\(\Rightarrow\frac{\left(11+x\right).n}{2}\)=11+10+...+x=0

=>11+x=0

=>x=-11

b/ \(\sqrt{12-\dfrac{12}{x^2}}+\sqrt{x^2-\dfrac{12}{x^2}}=x^2\)

\(\Leftrightarrow x-\sqrt{12-\dfrac{12}{x^2}}=\sqrt{x^2-\dfrac{12}{x^2}}\)

Bình phương 2 vế rút gọn

\(\Leftrightarrow x^4-x^2-4\sqrt{3\left(x^4-x^2\right)}+12=0\)

Đặt \(\sqrt{x^4-x^2}=a\)

\(\Rightarrow a^2-4\sqrt{3}a+12=0\)

\(\Leftrightarrow a=2\sqrt{3}\)

\(\Leftrightarrow x^4-x^2=12\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Câu a xem lại đề đúng không b. Do nghiệm xấu lắm

\(=tan^2\frac{\pi}{12}+tan^2\left(\frac{\pi}{2}-\frac{\pi}{12}\right)=tan^2\frac{\pi}{12}+cot^2\frac{\pi}{12}\)

\(=tan^2\frac{\pi}{12}+cot^2\frac{\pi}{12}+2tan\frac{\pi}{12}.cot\frac{\pi}{12}-2\)

\(=\left(tan\frac{\pi}{12}+cot\frac{\pi}{12}\right)^2-2\)

\(=\left(\frac{1}{sin\frac{\pi}{12}.cos\frac{\pi}{12}}\right)^2-2\)

\(=\left(\frac{2}{2sin\frac{\pi}{12}.cos\frac{\pi}{12}}\right)^2-2\)

\(=\left(\frac{2}{sin\frac{\pi}{6}}\right)^2-2\)

\(=16-2=14\)

cảm ơn bn :))