\(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2xz\) Thay x+y+z=0 vào

\(\Rightarrow0=x^2+y^2+z^2+2\left(xy+yz+xz\right)\)

\(\Leftrightarrow x^2+y^2+z^2=-2\left(xy+yz+xz\right)\) (1)

Ta có

\(\left(x^2+y^2+z^2\right)^2=x^4+y^4+z^4+2x^2y^2+2y^2z^2+2x^2z^2\) (2)

Bình phương 2 vế của (1)

\(\left(x^2+y^2+z^2\right)^2=4\left(xy+yz+xz\right)^2\)

\(\Leftrightarrow\left(x^2+y^2+z^2\right)^2=4\left(x^2y^2+y^2z^2+x^2z^2+2xy^2z+2xyz^2+2x^2yz\right)\)

\(\Leftrightarrow\left(x^2+y^2+z^2\right)^2=4\left[x^2y^2+y^2z^2+x^2z^2+2xyz\left(x+y+z\right)\right]\)

Do x+y+z=0 nên

\(\left(x^2+y^2+z^2\right)^2=4\left(x^2y^2+y^2z^2+x^2z^2\right)\)

\(\Rightarrow\dfrac{\left(x^2+y^2+z^2\right)^2}{2}=2x^2y^2+2y^2z^2+2x^2z^2\) (3)

Thay (3) vào (2)

\(\left(x^2+y^2+z^2\right)^2=x^4+y^4+z^4+\dfrac{\left(x^2+y^2+z^2\right)^2}{2}\)

\(\Rightarrow2\left(x^4+y^4+z^4\right)=\left(x^2+y^2+z^2\right)^2\) (đpcm)

Bạn xem lại đề

A C B H c b b' c'

Xét tg vuông ABC

\(c=\sqrt{a^2-b^2}=\sqrt{25^2-15^2}=\sqrt{400}=20cm\) (pitago)

\(b^2=b'.a\) (Trong tg vuông bình phương 1 cạnh góc vuông bằng tích giữa hình chiếu cạnh góc vuông đó trên cạnh huyền với cạnh huyền)

\(\Rightarrow b'=\dfrac{b^2}{a}=\dfrac{15^2}{25}=9cm\)

\(c'=a-b'=25-9=16cm\)

\(h^2=b'.c'\) (trong tg vuông bình phương đường cao hạ từ đỉnh góc vuông xuống cạnh huyền bằng tích giữa hình chiếu 2 cạnh góc vuông trên cạnh huyền)

\(\Rightarrow h=\sqrt{b'.c'}=\sqrt{15.9}=12cm\)

Hệ thức lượng trong tam giác vuông :

\(a^2=b^2+c^2\Rightarrow c^2=a^2-b^2=25^2-15^2=400\Rightarrow c=20\left(cm\right)\) 

\(b^2=a.b'\Rightarrow b'=b^2:a=15^2:25=9\left(cm\right)\)

\(c^2=a.c'\Rightarrow c'=c^2:a=25^2:25=25\left(cm\right)\)

\(h^2=b'.c'=9.25=225\Rightarrow h=15\left(cm\right)\)

     2\(\sqrt{\dfrac{16}{3}}\)  - 3\(\sqrt{\dfrac{1}{27}}\) - \(\dfrac{3}{2\sqrt{3}}\)

= \(\dfrac{8}{\sqrt{3}}\) - \(\dfrac{3}{3\sqrt{3}}\)  - \(\dfrac{3}{2\sqrt{3}}\)

= \(\dfrac{8}{\sqrt{3}}\) - \(\dfrac{1}{\sqrt{3}}\) - \(\dfrac{3}{2\sqrt{3}}\)

= \(\dfrac{16}{2\sqrt{3}}\) - \(\dfrac{2}{2\sqrt{3}}\) - \(\dfrac{3}{2\sqrt{3}}\)

= \(\dfrac{11}{2\sqrt{3}}\)

= \(\dfrac{11\sqrt{3}}{6}\)

f, 2\(\sqrt{\dfrac{1}{2}}\)- \(\dfrac{2}{\sqrt{2}}\) + \(\dfrac{5}{2\sqrt{2}}\)

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

= \(\dfrac{5}{2\sqrt{2}}\)

= \(\dfrac{5\sqrt{2}}{4}\)

(1 + \(\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\)).(1- \(\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\)) 

= \(\dfrac{\sqrt{3}-1+3-\sqrt{3}}{\sqrt{3}-1}\).\(\dfrac{\sqrt{3}+1-3+\sqrt{3}}{\sqrt{3}+1}\)

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

= \(\dfrac{-4}{3-1}\)

= \(\dfrac{-4}{2}\)

= -2

Bạn nên chịu khó gõ đề ra khả năng được giúp sẽ cao hơn.

Câu h của em đây nhé

h, ( 1 + \(\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\)).(1 - \(\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\))

= \(\dfrac{\sqrt{3}-1+3-\sqrt{3}}{\sqrt{3}-1}\).\(\dfrac{\sqrt{3}+1-3-\sqrt{3}}{\sqrt{3}+1}\)

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

= \(\dfrac{-4}{2}\)

= -2

Đề không đầy đủ. Bạn xem lại.

Hệ thức lượng trong tam giác vuông :

\(AB^2=BC.BH\left(1\right)\)

\(AC^2=BC.CH\left(2\right)\)

\(\left(1\right):\left(2\right)\Rightarrow\dfrac{BH}{CH}=\dfrac{AB^2}{AC^2}=\dfrac{25}{36}\left(\dfrac{AB}{AC}=\dfrac{5}{6}\right)\)

\(\Rightarrow BH=\dfrac{25}{36}CH\)

mà \(AH^2=BH.CH\)

\(\Rightarrow\dfrac{25}{36}CH^2=AH^2=30^2\)

\(\Rightarrow\dfrac{5}{6}CH=30\Rightarrow CH=\dfrac{30.6}{5}=36\) (\(\left(cm\right)\)

\(\Rightarrow BH=\dfrac{25}{36}.36=25\) \(\left(cm\right)\)

A B C H

Xét tg vuông ABH và tg vuông ACH có

\(\widehat{BAH}=\widehat{ACH}\) (cùng phụ với \(\widehat{ABC}\) )

=> tg ABH đồng dạng với tg ACH

\(\Rightarrow\dfrac{AH}{HC}=\dfrac{HB}{AH}=\dfrac{AB}{AC}=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{30}{HC}=\dfrac{5}{6}\Rightarrow HC=\dfrac{6.30}{5}=36cm\)

\(\Rightarrow\dfrac{HB}{30}=\dfrac{5}{6}\Rightarrow HB=\dfrac{5.30}{6}=25cm\)

In spite of = Due to + Ving/N, S + V

= In spite of/Due to + the fact that + S + V, S + V

=Although + S + V, S + V

Nghĩa: Mặc dù, biểu thị 2 hành động trái ngược nhau

a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b, Ta có: \(n_{Fe}=\dfrac{19,6}{56}=0,35\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Fe}=0,35\left(mol\right)\Rightarrow V_{H_2}=0,35.22,4=7,84\left(l\right)\)

c, \(n_{H_2SO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)

d, \(n_{FeSO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow m_{FeSO_4}=0,35.152=53,2\left(g\right)\)

e, \(C_{M_{FeSO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)

d, \(n_{H_2SO_4}=0,25.1,6=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{n_{Fe}}{1}< \dfrac{n_{H_2SO_4}}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{Fe}=0,35\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,4-0,35=0,05\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05.98=4,9\left(g\right)\)

\(\sqrt{\left(4-3\sqrt{2}\right)^2}=\left|4-3\sqrt{2}\right|=3\sqrt{2}-4\)

\(\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2+\sqrt{5}\right|=2+\sqrt{5}\\ \sqrt{\left(4+\sqrt{2}\right)^2}=\left|4+\sqrt{2}\right|=4+\sqrt{2}\)

\(\sqrt{6-2\sqrt{5}}=\sqrt{\sqrt{5^2}-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\\ \sqrt{7+4\sqrt{3}}=\sqrt{\sqrt{3^2}+2.2\sqrt{3}+2^2}=\sqrt{\left(\sqrt{3}+2\right)^2}=\left|\sqrt{3}+2\right|=\sqrt{3}+2\\ \sqrt{12-6\sqrt{3}}=\sqrt{\sqrt{3^2}-2.3\sqrt{3}+3^2}=\sqrt{\left(\sqrt{3}-3\right)^2}=\left|\sqrt{3}-3\right|=3-\sqrt{3}\)

\(\sqrt{17+12\sqrt{2}}=\sqrt{\left(2\sqrt{2}\right)^2+2.2\sqrt{2}.3+3^2}=\sqrt{\left(2\sqrt{2}+3\right)^2}=\left|2\sqrt{2}+3\right|=2\sqrt{2}+3\)

\(\dfrac{\sqrt{2}-\sqrt{11+6\sqrt{2}}}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\\ =\dfrac{\sqrt{2}-\sqrt{\sqrt{2^2}+2.3\sqrt{2}+3^2}}{\sqrt{\sqrt{5^2}+2\sqrt{5}+1}-\sqrt{5}}\\ =\dfrac{\sqrt{2}-\sqrt{\left(\sqrt{2}+3\right)^2}}{\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{5}}\\ =\dfrac{\sqrt{2}-\left|\sqrt{2}+3\right|}{\left|\sqrt{5}+1\right|-\sqrt{5}}\\ =\dfrac{\sqrt{2}-\sqrt{2}-3}{\sqrt{5}+1-\sqrt{5}}\\ =-3\)

\(\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{6+2\left|\sqrt{3}-1\right|}=\sqrt{6+2\sqrt{3}-2}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)

-3