a) \(\dfrac{2x}{3}\)+\(\dfrac{2x-1}{6}\)=4 - \(\dfrac{x}{3}\)

<=>\(\dfrac{2x}{3}\)+\(\dfrac{2x-1}{6}\) - 4+\(\dfrac{x}{3}\)=0

<=>\(\dfrac{2x.2+2x-1-4.6+x.2}{6}\)=0

=>4x-2x-24+2x=0

<=>4x-24=0

<=>4x=24

<=>x=6

Vậy x=6

b)\(\dfrac{x-1}{2}\)+\(\dfrac{x-1}{4}\)=1 - \(\dfrac{2\left(x-1\right)}{3}\)

<=>\(\dfrac{x-1}{2}\)+\(\dfrac{x-1}{4}\)-1+\(\dfrac{2\left(x-1\right)}{3}\)=0

<=>\(\dfrac{6.\left(x-1\right)+3\left(x-1\right)-1.12+4.2\left(x-1\right)}{12}\)=0

=>6x-6+3x-3-12+4x-4+2x-2=0

<=>15x-27=0

<=>15x=27

<=>x=\(\dfrac{9}{5}\)

Vậy x=\(\dfrac{9}{5}\)

d)\(\dfrac{7x}{8}\)-5(x-9)=20x+\(\dfrac{1,5}{6}\)

<=>\(\dfrac{7x}{8}\)-5(x-9)-20x-\(\dfrac{1,5}{6}\)=0

<=>\(\dfrac{3.7x-24.5\left(x-9\right)-20x.24-1,5.4}{24}\)=0

=>-478x+165=0

<=> - 478= - 165

<=> x=\(\dfrac{165}{487}\)

Vậy x =\(\dfrac{165}{487}\)

\(\sqrt{20}-\sqrt{45}+\sqrt{6+2\sqrt{5}}=\sqrt{2^2.5}-\sqrt{3^2.5}+\sqrt{\left(\sqrt{5}+1\right)^2}=2\sqrt{5}-3\sqrt{5}+\sqrt{5}+1=1\)

\(\sqrt{20}-2-\sqrt{\left(\sqrt{5}-2\right)^2}=2\sqrt{5}-2-\left|\sqrt{5}-2\right|=2\sqrt{5}-2-\sqrt{5}+2=\sqrt{5}\)

\(\left(\sqrt{27}+3\sqrt{12}-2\sqrt{3}\right):\sqrt{3}=\left(3\sqrt{3}+6\sqrt{3}-2\sqrt{3}\right):\sqrt{3}=7\sqrt{3}:\sqrt{3}=7\)

\(\sqrt{50}-3\sqrt{8}+\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{5^2.2}-3\sqrt{2^2.2}+\sqrt{\left(\sqrt{2}+1\right)^2}=5\sqrt{2}-6\sqrt{2}+\sqrt{2}+1=1\)

1) \(A=\sqrt{20}-\sqrt{45}+\sqrt{6+2\sqrt{5}}\)

\(=2\sqrt{5}-3\sqrt{5}+\sqrt{5}+1\)

=1

2) Ta có: \(B=\sqrt{20}-2-\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(=2\sqrt{5}-2-\sqrt{5}+2\)

\(=\sqrt{5}\)

3) Ta có: \(\left(\sqrt{27}+3\sqrt{12}-2\sqrt{3}\right):\sqrt{3}\)

\(=\left(3\sqrt{3}+6\sqrt{3}-2\sqrt{3}\right):\sqrt{3}\)

=7

4) Ta có: \(\sqrt{50}-3\sqrt{8}+\sqrt{\left(\sqrt{2}+1\right)^2}\)

\(=5\sqrt{2}-6\sqrt{2}+\sqrt{2}+1\)

=1

1.

a.

ĐKXĐ: \(x^2-1>0\Rightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)

\(log_2\left(x^2-1\right)=3\)

\(\Rightarrow x^2-1=8\)

\(\Leftrightarrow x^2=9\)

\(\Rightarrow x=\pm3\) (tm)

b.

ĐKXĐ: \(x>0\)

\(log_3x+log_{\sqrt{3}}x+log_{\dfrac{1}{3}}x=6\)

\(\Leftrightarrow log_3x+2log_3x-log_3x=6\)

\(\Leftrightarrow log_3x=3\)

\(\Rightarrow x=3^3=27\)

c. ĐKXĐ: \(x>0\)

\(log_{\sqrt{2}}^2x+3log_2x+log_{\dfrac{1}{2}}x=2\)

\(\Leftrightarrow\left(2log_2x\right)^2+3log_2x-log_2x=2\)

\(\Leftrightarrow4log_2^2x+2log_2x-2=0\)

\(\Rightarrow\left[{}\begin{matrix}log_2x=-1\\log_2x=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\sqrt{2}\end{matrix}\right.\)

d.

ĐKXĐ: \(x>0\)

\(log_{\dfrac{1}{2}}^24x+log_2\dfrac{x^2}{8}=8\)

\(\Leftrightarrow\left(-log_24x\right)^2+log_2x^2-log_28=8\)

\(\Leftrightarrow\left(log_2x+log_24\right)^2+2log_2x-3=8\)

\(\Leftrightarrow\left(log_2x+2\right)^2+2log_2x-11=0\)

\(\Leftrightarrow log_2^2x+6log_2x-7=0\)

\(\Rightarrow\left[{}\begin{matrix}log_2x=1\\log_2x=-7\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2^7}\end{matrix}\right.\)

Bài 1:

Áp dụng HTL trong tam giác vuông:
$AH^2=BH.CH$

$\Leftrightarrow x^2=4.9=36$

$\Rightarrow x=6$ (do $x>0$)

Bài 2:

Áp dụng định lý Pitago:

$BC=\sqrt{AB^2+AC^2}=\sqrt{6^2+8^2}=10$ (cm)

$\sin B=\frac{AC}{BC}=\frac{6}{10}=\frac{3}{5}$

$\Rightarrow \widehat{B}=36,87^0$

$\widehat{C}=90^0-\widehat{B}=90^0-36,87^0=53,13^0$

Bài 3:

a. Áp dụng định lý Pitago:

$BC=\sqrt{AB^2+AC^2}=\sqrt{3^2+4^2}=5$ (cm)

$AH=\frac{2S_{ABC}}{BC}=\frac{AB.AC}{BC}=\frac{3.4}{5}=2,4$ (cm)

b.

$\sin B=\frac{AC}{BC}=\frac{4}{5}$

$\Rightarrow \widehat{B}=53,13^0$

$\widehat{C}=90^0-\widehat{B}=36,87^0$

c.

Áp dụng tính chất tia phân giác:

$\frac{BE}{CE}=\frac{AB}{AC}=\frac{3}{4}$

$\Rightarrow \frac{BE}{BC}=\frac{3}{7}$

$\Rightarrow BE=BC.\frac{3}{7}=\frac{5.3}{7}=\frac{15}{7}$  (cm)

$CE=BC-BE=5-\frac{15}{7}=\frac{20}{7}$ (cm)

hoạt động dân gian ngày tết có ông địa cầm quạt phe phẩy ?

tôm khô ăn với củ gì vào ngày tết

món này ngày tết,năm quả trên mâm,nhà nào cũng có,vào khắc giao thừa

Hai câu đầu có đc gọi là câu đối không ?

có mà , mình tìm ở nhà mình 

2^x.4^12=8^9