\(x^2-4x-1=0\)

\(\left(x^2-2\cdot x\cdot2+4\right)-5=0\)

\(\left(x-2\right)^2=\left(\sqrt{5}\right)^2\)

\(\Rightarrow x-2=\pm\sqrt{5}\)

Tự giải tiếp nha ...

\(x^2-4x-1=0\)

\(\Delta=\left(-4\right)^2-4.\left(-1\right)=20\)

pt có 2 nghiệm

\(x_1=\frac{-4-\sqrt{20}}{2}=-2-\sqrt{5}\)

\(x_2=\frac{-4+\sqrt{20}}{2}=-2+\sqrt{5}\)

Bài 3: 

PTHH: \(2R+O_2\xrightarrow[]{t^O}2RO\)

Theo PTHH: \(n_R=n_{RO}\) \(\Rightarrow\dfrac{3,6}{M_R}=\dfrac{6}{M_R+16}\)

  \(\Rightarrow M_R=24\)  (Magie)

Bài 4 ở đâu vậy ??

Ta có : \(\left(3x-2\right)\left(4x+3\right)=\left(2-3x\right)\left(x-1\right)\)

\(\Leftrightarrow12x^2-8x+9x-6=2x-3x^2-2+3x\)

\(\Leftrightarrow12x^2-8x+9x-6-2x+3x^2+2-3x=0\)

\(\Leftrightarrow15x^2-4x-4=0\)​

\(\Leftrightarrow15x^2-10x+6x-4=0\)

Lỗi :vvvv

\(\Leftrightarrow10x\left(\dfrac{3}{2}x-1\right)+4\left(\dfrac{3}{2}x-1\right)=0\)

\(\Leftrightarrow\left(10x+4\right)\left(\dfrac{3}{2}x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy ...

Bài 5: 

a: BC=10cm

b: HA=4,8cm

HB=3,6(cm)

HC=6,4(cm)

Bài 6:

\(x^3=6+3\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\\ \Leftrightarrow x^3=6+3x\sqrt[3]{1}\\ \Leftrightarrow x^3-3x=6\\ y^3=34+3\sqrt[3]{\left(17+12\sqrt{2}\right)\left(17-12\sqrt{2}\right)}\left(\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}}\right)\\ \Leftrightarrow y^3=34+3y\sqrt[3]{1}\\ \Leftrightarrow y^3-3y=34\\ \Leftrightarrow P=x^3-3x+y^3-3y+1980=6+34+1980=2020\)

a, Ta có : \(P\left(x\right)+Q\left(x\right)\)hay 

\(3x^5-4x^4+2x^3-7x+1+x^5-x^3+4x-5=4x^5-4x^4+x^3-3x-4\)

b, Ta có : \(P\left(x\right)-Q\left(x\right)\)hay 

\(3x^5-4x^4+2x^3-7x+1-x^5+x^3-4x+5=2x^5-4x^4+3x^3-11x+6\)

BẠN GIÚP MÌNH TÍNH THEO CÁCH 1 VÀ 2 TRONG SGK VỚI

ở chỗ 19?16 là em viết sai thật là 19/16 ạ

Mọi người trả lời nhanh giúp em với ạ

\(\frac{2}{x-2}-\frac{3}{x+2}=\frac{x+1}{x^2-4}\left(x\ne\pm2\right)\)

\(\Leftrightarrow\frac{2}{x-2}-\frac{3}{x+2}-\frac{x+1}{x^2-4}=0\)

\(\Leftrightarrow\frac{2}{x-2}-\frac{3}{x+2}-\frac{x+1}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{3\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x+1}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{2x+4-3x+6-x-1}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{-2x-9}{\left(x-2\right)\left(x+2\right)}=0\)

=> -2x-9=0

<=> -2x=9

<=> \(x=\frac{-9}{2}\left(tmđk\right)\)

=\(\left(2-1\right)\left(2+1\right)\left(2^2-1\right)....\left(2^{20}-1\right)\) +1

=\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{20}+1\right)+1\)

=\(\left(2^4-1\right)\left(2^4+1\right)....\left(2^{20}+1\right)+1\)

=.....

=\(\left(2^{20}-1\right)\left(2^{20}+1\right)+1\)

=\(2^{40}-1+1\)

=\(2^{40}\)

Chuc ban hoc tot

Sai rồi, nếu mũ là 32 thì bài này làm thế đc chứ mũ 20 thì ko làm như này được