\(\sqrt{x}+1=40\Rightarrow\sqrt{x}=39\Rightarrow\left(\sqrt{x}\right)^2=39^2\Rightarrow x=1521\)

\(3\sqrt{x}+1=40\)

ĐK : x ≥ 0

<=> \(3\sqrt{x}=39\)

<=> \(\sqrt{x}=13\)

<=> \(x=169\)( tm )

Vậy x = 169

a) Ta có \(\left|x-4\right|\ge0\forall x\Rightarrow A=7+\left|x-4\right|\ge7\forall x\)

Dấu "=" xảy ra <=> x - 4 = 0

=> x = 4

Vậy Min A = 7 <=> x = 4

b) Ta có : \(\left|2-3x\right|\ge0\forall x\Rightarrow B=\left|2-3x\right|-\frac{1}{5}\ge-\frac{1}{5}\forall x\)

Dấu "=" xảy ra <=> 2 - 3x = 0

=> 3x = 2

=> x = 2/3

Vậy Min B = -1/5 <=> x = 2/3

c) Ta có \(\left|\frac{1}{2}-5x\right|\ge0\forall x\Rightarrow C=7-\left|\frac{1}{2}-5x\right|\le7\forall x\)

Dấu "=" xảy ra <=> 1/2 - 5x = 0

=> x = 1/10 

Vậy Max C = 7 <=> x = 1/10

Ta có: \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

Suy ra:

 \(\frac{a}{b+c}=\frac{1}{2}\Rightarrow a=\frac{b+c}{2}=\frac{1}{2}\times\left(b+c\right)\)

\(\frac{b}{a+c}=\frac{1}{2}\Rightarrow b=\frac{a+c}{2}=\frac{1}{2}\times\left(a+c\right)\)

\(\frac{c}{a+b}=\frac{1}{2}\Rightarrow c=\frac{a+b}{2}=\frac{1}{2}\times\left(a+b\right)\)

Thay  \(a=\frac{1}{2}\times\left(b+c\right)\);  \(b=\frac{1}{2}\times\left(a+c\right)\); \(c=\frac{1}{2}\times\left(a+b\right)\) vào P ta được:

\(\frac{b+c}{\frac{1}{2}\times\left(b+c\right)}+\frac{c+a}{\frac{1}{2}\times\left(a+c\right)}+\frac{a+b}{\frac{1}{2}\times\left(a+b\right)}\)

\(=\frac{\text{ }1\text{ }}{\frac{1}{2}}+\frac{1}{\frac{1}{2}}+\frac{1}{\frac{1}{2}}\)

\(=2+2+2=6\)

Vậy giá trị của P  là 6

a, \(\left|x-\frac{2}{3}\right|=\frac{1}{2}\)

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

b, \(\left|x+\frac{7}{20}\right|=\frac{3}{15}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{7}{20}=\frac{1}{5}\\x+\frac{7}{20}=-\frac{1}{5}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}-\frac{7}{20}\\x=-\frac{1}{5}-\frac{7}{20}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{-3}{20}\\x=\frac{-11}{20}\end{cases}}}\)

c, \(\left|3x+2\right|=\left|7x-4\right|\)

\(\Leftrightarrow\orbr{\begin{cases}3x+2=7-4x\\3x+2=4x-7\end{cases}\Leftrightarrow\orbr{\begin{cases}7x=5\\x=9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{7}\\x=9\end{cases}}}\)

d, \(\left|5-2x\right|=\left|2x-5\right|\)

\(\Leftrightarrow\orbr{\begin{cases}5-2x=2x-5\\5-2x=5-2x\end{cases}\Leftrightarrow\orbr{\begin{cases}-4x=-10\\0x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x\in Q\end{cases}}}\)

=> Có vô số x thỏa mãn \(\left|5-2x\right|=\left|2x-5\right|\)

e, \(\left|-5-6x\right|=\left|-x-5\right|\)

\(\Leftrightarrow\orbr{\begin{cases}-5-6x=-x-5\\-5-6x=x+5\end{cases}\Leftrightarrow\orbr{\begin{cases}-5x=0\\-7x=10\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=-\frac{10}{7}\end{cases}}}\)

f, \(\left|-x+5\right|=\left|12-3x\right|\) đúng ko ???

\(\Leftrightarrow\orbr{\begin{cases}-x-5=12-3x\\-x+5=3x-12\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\-4x=17\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{17}{4}\end{cases}}}\)

a/

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(A=2A-A=1-\frac{1}{2^{100}}< 1\)

b/

\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\)

\(2B=3B-B=1-\frac{1}{3^{2019}}\Rightarrow B=\frac{1}{2}-\frac{1}{2.3^{2019}}< \frac{1}{2}\)

Phần C đề thiếu

\(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)

\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)

\(\Rightarrow3D-D=(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}})-\)\((\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}})\)

\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(\Rightarrow6D=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow6D-2D=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}\)

\(\Rightarrow4D=3-\frac{203}{3^{100}}\)

\(\Rightarrow D=\frac{3}{4}-\frac{\frac{203}{3^{100}}}{4}< \frac{3}{4}\left(đpcm\right)\)

sửa rồi nhá bn

\(M=\frac{2012}{2013}.\frac{2012^{2011}}{2013^{2011}}\)

\(N=\frac{2012}{2013}.\frac{2012^{2011}+1}{2013^{2011}+1}\)

Bạn tự so sánh tiếp nhé!

Đặt 2012²⁰¹² = x ; 2013²⁰¹³ = y

Giả sử M < N 

Ta có : \(\frac{x}{y}< \frac{x+2012}{y+2013}\)

\(\Leftrightarrow x\left(y+2013\right)< y\left(x+2012\right)\)

\(\Leftrightarrow xy+2013x< xy+2012y\)

\(\Leftrightarrow2013x< 2012y\)

\(\Leftrightarrow2013.2012^{2012}< 2012.2013^{2013}\)

\(\Leftrightarrow2012^{2011}< 2013^{2012}\)( Đúng )

=> Điều giả sử trên là đúng

=> M < N