Bạn ơi ở mẫu kia là dấu j vây sao lại "="

a)\(\left(a-m\right)-\left(m-b+c\right)+\left(-b-c\right)=a-m-m+b-c-b-c=a-2c-2m\)

b) \(\left(a+b-c\right)-\left(-c+b\right)+a=a+b-c+c-b+a=2a\)

c)\(\left(-a+b-c\right)-\left(-a-b-c\right)=-a+b-c+a+b+c=2b\)

d)\(\left(b+d\right)+\left(-b-c\right)-\left(c-a\right)-a=b+d-b-c-c+a=a-2c+d\)

a)Có: \(\sqrt{2}\cdot\sqrt{7+3\sqrt{5}}\)

       \(=\sqrt{14+6\sqrt{5}}=\sqrt{9+2\cdot3\cdot\sqrt{5}+5}=\sqrt{\left(3+\sqrt{5}^2\right)}=3+\sqrt{5}\)

=> \(\sqrt{7+3\sqrt{5}}=\frac{3+\sqrt{5}}{\sqrt{2}}\)

b)\(\sqrt{118+28\sqrt{10}}\)

\(=\sqrt{2\left(59+14\sqrt{10}\right)}\)

\(=\sqrt{2\left(49+2\cdot7\cdot\sqrt{10}+10\right)}\)

\(=\sqrt{2\left(7+\sqrt{10}\right)^2}\)

\(=\sqrt{2}\left(7+\sqrt{10}\right)\)

a) \(5\cdot\left(\frac{x}{3}-4\right)=15\)

\(\Leftrightarrow\)\(\frac{x-12}{3}=3\)

\(\Leftrightarrow x-12=9\)

\(\Leftrightarrow x=21\)

Vạy x=21

+) 2x+3 chia hét cho x+1

Bạn chia cột dọc 2x+3 : x+1 =2 dư 1

Vậy để 2x+3 \(⋮\) x+1 thì x+1 \(\in\) Ư(1)

Mà Ư(1)={1;-1}

=> x+1={1;-1}

*)TH1: x+1=1<=>x=0

*)TH2: x+1=-1<=>x=-2

Vậy x={-2;0} thì 2x+3\(⋮\) x+1

b)Tìm GTLN của \(\frac{7}{\left(x+1\right)^2+1}\)

Vì \(\left(x+1\right)^2\ge0\) với mọi x

=>\(\left(x+1\right)^2+1\ge1\) 

=> \(\frac{7}{\left(x+1\right)^2+1}\le\frac{7}{1}=7\)

a) Vì BD là tia pg giác của \(\widehat{ABC}\) (gt)

=>\(\frac{AB}{BC}=\frac{AD}{DC}\)

=>\(\frac{AB}{AB+AC}=\frac{AD}{AD+DC}\)

=> \(\frac{AB}{AB+BC}=\frac{AD}{AC}\)

=>\(\frac{20}{20+5}=\frac{AD}{20}\)

=>\(AD=\frac{20\cdot20}{20+5}=16\) cm

Có: AC=AD+DC 

=>DC=AC-AD=20-16=4 cm

\(x^4+2014x^2+2013x+2014\)

\(=x^4+2014x^2+2014x-x+2014\)

\(=\left(x^4-x\right)+\left(2014x^2+2014x+2014\right)\)

\(=x\left(x^3-1\right)+2014\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2014\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2014\right)\)

b)\(x^8+7x^4+6\)

\(=x^8+x^4+6x^4+6\)

\(=x^4\left(x^4+1\right)+6\left(x^4+1\right)\)

\(=\left(x^4+1\right)\left(x^4+6\right)\)

\(\left(x-3\right)\left(x+1\right)>0\)

\(\Leftrightarrow\begin{cases}x-3>0\\x+1>0\end{cases}\) hoặc \(\begin{cases}x-3< 0\\x+1< 0\end{cases}\)

\(\Leftrightarrow\begin{cases}x>3\\x>-1\end{cases}\) hoặc \(\begin{cases}x< 3\\x< -1\end{cases}\)

\(\Leftrightarrow x>3\) hoặc \(x< -1\)

\(M=\frac{7}{\left(x+1\right)^2+1}\)

Vì \(\left(x+1\right)^2\ge0\) Với mọi x

=> \(\left(x+1\right)^2+1\ge1\)

=>\(\frac{7}{\left(x+1\right)^2+1}\le\frac{7}{1}=7\)

Vậy GTLN của M là 7 khi x=-1

\(\Delta\)ABC có: DA=DB(gt)  

                 EA=EC(gt)

=> DE là đường trung bình của \(\Delta\)ABC

=> DE//BC

Xét tứ giác BDEC có: DE//BC

=> Tứ giác BDEC là hình thang

Mà:^B=^C (gt)

=> Tứ giác BDEC là hình thang cân

b)Vì DE là đường trung bình của tam giác ABC

=>\(DE=\frac{1}{2}BC=\frac{1}{2}\cdot8=4\)

a)ĐK:\(\begin{cases}x^2-1\ge0\\x^2-2\sqrt{x^2-1}\ge0\end{cases}\)\(\Leftrightarrow\begin{cases}x^2\ge1\\x^2\ge2\sqrt{x^2-1}\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge1\\x^4\ge4\left(x^2-1\right)\end{cases}\)

\(\Leftrightarrow\begin{cases}x\ge1\\x^4-4x^2+4\ge0\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge1\\\left(x^2-2\right)^2\ge0\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge1\\x^2-2\ge0\end{cases}\)

\(\Leftrightarrow\begin{cases}x\ge1\\x^2\ge2\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge1\\x\ge\sqrt{2}\end{cases}\)\(\Leftrightarrow x\ge\sqrt{2}\)

b)Có \(A=\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\)

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

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

\(=\sqrt{x^2-1}+1-\left|\sqrt{x^2-1}-1\right|\)

Vói \(x\ge1\) thì A=\(\sqrt{x^2-1}+1-\left(\sqrt{x^2-1}-1\right)=\sqrt{x^2-1}+1-\sqrt{x^2-1}+1=2\)

Với \(\sqrt{2}< x< 1\) thì 

                \(A=\sqrt{x^2-1}+1-\left(1-\sqrt{x^2-1}\right)=\sqrt{x^2-1}+1-1+\sqrt{x^2-1}=2\sqrt{x^2-1}\)