a, \(\sqrt{4-5x}=12\Leftrightarrow4-5x=144\Leftrightarrow5x=140\Leftrightarrow x=28\)

b,ĐK :  \(x\ge7\)

 \(\sqrt{x^2-14x+49}-3x=1\Leftrightarrow\sqrt{\left(x-7\right)^2}=3x+1\)

\(\Leftrightarrow x-7=3x+1\Leftrightarrow-2x-8=0\Leftrightarrow x=-4\)( vô lí )

c, Bn làm nốt nhé 

a) đk: \(x\le\frac{4}{5}\)

Ta có: \(\sqrt{4-5x}=12\)

\(\Leftrightarrow\left|4-5x\right|=144\)

\(\Rightarrow4-5x=144\)

\(\Leftrightarrow5x=-140\)

\(\Rightarrow x=-28\left(tm\right)\)

b) Ta có: \(\sqrt{x^2-14x+49}-3x=1\)

\(\Leftrightarrow\sqrt{\left(x-7\right)^2}=1+3x\)

\(\Leftrightarrow\left|x-7\right|=3x+1\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=3x+1\\x-7=-3x-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-8\\4x=6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=\frac{3}{2}\end{cases}}\)

Ấn CTRL + f5 nhé

không có ảnh cho sẵn à

a) Gọi A = 1/2+1/2^2+1/2^3+1/2^4+...+1/2^n

Có 2A = 1+ 1/2+ 1/2^2 +1/2^3 + ... + 1/2^n-1

2A-A= 1 - 1/2^n

=> A = 1 - 1/2^n

Có 1 - 1/2^n < 1

=> A<1

Ta có : \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{10.10}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}=\frac{9}{10}< 1\left(\text{đpcm}\right)\)

đề hơi sai sai nha bạn , xem lại nhé

cậu viết sai đề bài rồi

1 pm đến 6 pm

1 pm - 6pm

Ta có \(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2020}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2019}{2020}=\frac{1.2.3...2019}{2.3.4...2020}=\frac{1}{2020}\)

Lại có : \(A=\left(1\frac{1}{2}\right).\left(1\frac{1}{3}\right).\left(1\frac{1}{4}\right)...\left(1\frac{1}{2020}\right)\)

\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{2021}{2020}=\frac{3.4.5...2021}{2.3.4...2020}=\frac{2021}{2}\)

Khi đó \(\frac{A}{B}=\frac{\frac{2021}{2}}{\frac{1}{2020}}=\frac{2021}{2}.2020=2041210\)

a) Ta có: \(\sqrt{16-6\sqrt{7}}+\sqrt{7}\)

\(=\sqrt{3^2-2.3.\sqrt{7}+7}+\sqrt{7}\)

\(=\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{7}\)

\(=\left|3-\sqrt{7}\right|+\sqrt{7}\)

\(=3-\sqrt{7}+\sqrt{7}\)

\(=3\)

b) Ta có: \(\sqrt{\left|12\sqrt{5}-29\right|}+\sqrt{12\sqrt{5}+29}\)

\(=\sqrt{\left(\sqrt{29-12\sqrt{5}}+\sqrt{12\sqrt{5}+29}\right)^2}\)

\(=\sqrt{29-12\sqrt{5}+2\sqrt{\left(29-12\sqrt{5}\right)\left(12\sqrt{5}+29\right)}+12\sqrt{5}+29}\)

\(=\sqrt{58+2\sqrt{121}}\)

\(=\sqrt{58+2.11}\)

\(=\sqrt{80}=4\sqrt{5}\)