\(\sqrt{2.\sqrt{4}}\)\(=\sqrt{2.2}\)\(=\sqrt{4}\)\(=2\)
\(23+223+2223+22223=24692\)
\(\frac{7}{5}:\frac{5}{4}=\frac{7}{5}.\frac{4}{5}=\frac{28}{25}\)
b, \(M=A-B=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\left(\frac{5}{x+\sqrt{x}-6}+\frac{1}{\sqrt{x}-2}\right)\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}-\frac{1}{\sqrt{x}-2}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{x+\sqrt{x}-6}-\frac{5}{x+\sqrt{x}-6}-\frac{1\left(\sqrt{x}+3\right)}{x+\sqrt{x}-6}\)
\(=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-4\sqrt{x}+3\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)\(=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)
CHỨNG MINH RẰNG VỚI MỌI SỐ DƯƠNG N THÌ
GIÚP MÌNH VỚI
1+\(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-1\right)\)
tính hộ chúa con cuối với " ko dùng coccoc math " 100% sai " bạn nào có máy tính casio bấm hộ "
\(x^2+3=x+8+2x-x^2+2x\sqrt{8+2x-x^2}.\)
\(2x^2-3x-5=2x\sqrt{8+2x-x^2}\)
\(4x^4-12x^3-11x^2+30x+25=-4x^4+8x^3+32x^2\)
\(\left(X+1\right)^2\left(2x-5\right)^2+4x^4-8x^3-32x^2=0\)
\(\left(X-1\right)\left(8x^3-12x^2-55x-25\right)=0\)
\(8x^3-12x^2-55x-25=0\)
\(\Delta=144+1320=1464>0\)
\(k=\frac{47520+3456+43200}{2\sqrt{1464^3}}=\frac{94176}{2\sqrt{1464^3}}=\frac{47088}{\sqrt{1464^3}}< 1\)
\(x1=\frac{2\sqrt{1464}cos\left(arccos\left(\frac{47088}{\sqrt{1464^3}}\right)-\frac{2pi}{3}\right)+12}{24}=?\)
x2=...
x3=......
Tính: \(1+1\frac{3}{4}+\frac{6}{5}-5+12,4=?\)
\(5+4+\sqrt{100}-\sqrt{25}+12,5=?\)
[TEX]\frac{x}{2} = \frac{y}{3} <=> \frac{x}{8} = \frac{y}{12}[/TEX]
[TEX]\frac{y}{4} = \frac{z}{5} <=> \frac{y}{12} = \frac{z}{15}[/TEX]
Suy ra:
[TEX]\frac{x}{8} = \frac{y}{12} = \frac{z}{15} [/TEX]
Mặt khác: [TEX]x+y+z=10 [/TEX]
Áp dụng tính chấmơẻ rộng của dãy tỉ số bằng nhau:
[TEX]\frac{x+y+z}{8+12+15} = \frac{10}{35} = \frac{2}{7} [/TEX]
[TEX]x= \frac{16}{7}[/TEX]
[TEX]y= \frac{24}{7}[/TEX]
[TEX]z= \frac{30}{7}[/TEX]
1) Tính:
a) \(\frac{3}{5}+\left(-\frac{1}{4}\right)\)
b) \(\left(-\frac{5}{18}\right)\left(-\frac{9}{10}\right)\)
c) \(4\frac{3}{5}:\frac{2}{5}\)
2) Tìm x:
a)\(\frac{12}{x}=\frac{3}{4}\)
b) \(x:\left(\frac{-1}{3}\right)^3=\left(\frac{-1}{3}\right)^2\)
c) \(\frac{-11}{12}.x+0,25=\frac{5}{6}\)
d) \(\left(x-1\right)^5=-32\)
3) Cho |m| = -3, tìm m:
4) Các cạnh của một tam giác có số đo tỉ lệ với các số 3; 4; 5. Tính cạnh của tam giác biết chu vi của nó là 13,2 cm
\(B=x-4\sqrt{x}+\frac{x+16}{\sqrt{x}+3}+10=x-4\sqrt{x}+4+\frac{4\left(\sqrt{x}+3\right)+x-4\sqrt{x}+4}{\sqrt{x}+3}+6\)
\(=\left(\sqrt{x}-2\right)^2+\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+3}+4+6\ge10\)Dấu = xảy ra tại x=4
D= \(\frac{x^2-2x+2000}{x^2}\)= 1-\(\frac{2x}{x^2}\)+\(\frac{2000}{x^2}\)
Đặt t= \(\frac{1}{2}\)=> D= 2000t2-2t+1 = (\(20\sqrt{5}t\))2-2.\(20\sqrt{5}t\).\(\frac{1}{20\sqrt{5}}\)+\(\left(\frac{1}{20\sqrt{5}}\right)^2\)\(-\left(\frac{1}{20\sqrt{5}}\right)^2\)+1
D= (\(20\sqrt{5}t\)-\(\frac{1}{20\sqrt{5}}\)) 2+\(\frac{1999}{2000}\)\(\ge\)\(\frac{1999}{2000}\)
Min D= \(\frac{1999}{2000}\)khi \(20\sqrt{5}t\)\(-\frac{1}{20\sqrt{5}}\)= 0 => t = \(\frac{1}{2000}\)=> \(\frac{1}{x}\)= \(\frac{1}{2000}\)=> x= 2000
\(\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a+b\right)}\)
\(VP=\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\le\frac{a+b}{2}\sqrt{2\left(a+b\right)}\)\(\Rightarrow\)\(VP^2\le\frac{\left(a+b\right)^3}{2}\) (1)
chứng minh bổ đề: \(VT^2=\left(\frac{\left(a+b\right)^2}{2}+\frac{a+b}{4}\right)^2\ge\frac{\left(a+b\right)^3}{2}\)
\(\Leftrightarrow\)\(\frac{\left(a+b\right)^4}{4}+\frac{\left(a+b\right)^2}{16}+\frac{\left(a+b\right)^3}{4}\ge\frac{\left(a+b\right)^3}{2}\)
\(\Leftrightarrow\)\(\left(a+b\right)^4+\frac{\left(a+b\right)^2}{4}\ge\left(a+b\right)^3\)
Có: \(\left(a+b\right)^4+\frac{\left(a+b\right)^2}{4}\ge2\sqrt{\frac{\left(a+b\right)^6}{4}}=\left(a+b\right)^3\)\(\Rightarrow\)\(VT^2\ge\frac{\left(a+b\right)^3}{2}\) (2)
(1) và (2) => \(VT^2\ge VP^2\) => \(VT\ge VP\) ( đpcm )
