AI giúp mình bài 16 từ 1-10 với
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31 tháng 10 2021
Tham khảo!
Mùa xuân, cây gạo gọi đến bao nhiêu là chim. Từ xa nhìn lại, cây gạo sừng sững như một tháp đèn khổng lồ. Hàng ngàn bông hoa là hàng ngàn ngọn lửa hồng. Hàng ngàn búp nõn là hàng ngàn ánh nến trong xanh. Tất cả đều lóng lánh lung linh trong nắng. Chào mào, sáo sậu, sáo đen, đàn đàn lũ lũ bay đi bay về. Chúng nó gọi nhau, trêu ghẹo nhau, trò chuyện ríu rít. Ngày hội mùa xuân đấy...
Hết mùa hoa, chim chóc cũng vãn. Cây gạo chấm dứt những ngày tưng bừng, ồn ã, lại trở về với dáng vẻ xanh mát, trầm tư. Cây đứng im cao lớn, làm tiêu cho những con cò cập bến và những đứa con về thăm quê mẹ.
NV
6
23 tháng 3 2021
9.
\(\Leftrightarrow a^2+a^2b^2+b^2+b^2c^2+c^2+c^2a^2\ge6abc\)
\(\Leftrightarrow\left(a^2-2abc+b^2c^2\right)+\left(b^2-2abc+c^2a^2\right)+\left(c^2-2abc+a^2b^2\right)\ge0\)
\(\Leftrightarrow\left(a-bc\right)^2+\left(b-ca\right)^2+\left(c-ab\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(0;0;0\right);\left(1;1;1\right);\left(1;-1;-1\right)\) và các hoán vị
10.
\(a^2+b^2+c^2=1\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=1+2\left(ab+bc+ca\right)\)
\(\Leftrightarrow\left(a+b+c\right)^2=1+2\left(ab+bc+ca\right)\)
\(\Rightarrow1+2\left(ab+bc+ca\right)\ge0\Rightarrow ab+bc+ca\ge-\dfrac{1}{2}\)
Lại có:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow ab+bc+ca\le1\)
23 tháng 3 2021
11.
Do \(a^2+b^2+c^2=1\Rightarrow\left\{{}\begin{matrix}\left|a\right|\le1\\\left|b\right|\le1\\\left|c\right|\le1\end{matrix}\right.\) \(\Rightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge0\)
Do đó:
\(abc+2\left(1+a+b+c+ab+bc+ca\right)\)
\(=1+a+b+c+ab+bc+ca+\left(1+a+b+c+ab+bc+ca+abc\right)\)
\(=\dfrac{1}{2}\left(a^2+b^2+c^2\right)+ab+bc+ca+a+b+c+\dfrac{1}{2}+\left(a+1\right)\left(b+1\right)\left(c+1\right)\)
\(=\dfrac{1}{2}\left(a+b+c\right)^2+\left(a+b+c\right)+\dfrac{1}{2}+\left(a+1\right)\left(b+1\right)\left(c+1\right)\)
\(=\dfrac{1}{2}\left(a+b+c+1\right)^2+\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge0\) (đpcm)
8 tháng 11 2017
Ta có:2A=\(2+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
2A-A=\(\left(2+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)
\(=2-\frac{1}{32}=\frac{63}{32}=A\)
LJ
8 tháng 11 2017
Ta có: \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(\Rightarrow A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}\)
\(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}\right)\)
\(\Rightarrow A=1-\frac{1}{2^5}=\frac{31}{32}\)
Vậy \(A=\frac{31}{32}\)
8 tháng 4 2023
1)\(\dfrac{2}{9}+\dfrac{-3}{4}+\dfrac{5}{30}\)
\(=\dfrac{2.20}{9.20}+\dfrac{-3.45}{4.45}+\dfrac{5.6}{30.6}\)
\(=\dfrac{40}{180}+\dfrac{-135}{180}+\dfrac{30}{180}\)
\(=\dfrac{40+\left(-135\right)+30}{180}\)
\(=\dfrac{-65}{180}\)
\(=\dfrac{-13}{36}\)
2)\(\dfrac{-7}{12}-\dfrac{11}{18}\)
\(=\dfrac{-7.3}{12.3}-\dfrac{11.2}{18.2}\)
\(=\dfrac{-21}{36}-\dfrac{22}{36}\)
\(=\dfrac{-21-22}{36}\)
\(=\dfrac{-43}{36}\)
3)\(\dfrac{7}{8}-\dfrac{-5}{16}\)
\(=\dfrac{7.2}{8.2}-\dfrac{-5}{16}\)
\(=\dfrac{14}{16}-\dfrac{-5}{16}\)
\(=\dfrac{14-\left(-5\right)}{16}\)
\(=\dfrac{19}{16}\)
4)\(\dfrac{3}{8}-\dfrac{-9}{10}-\dfrac{5}{16}\)
\(=\dfrac{3.10}{8.10}-\dfrac{-9.8}{10.8}-\dfrac{5.5}{16.5}\)
\(=\dfrac{30}{80}-\dfrac{-72}{80}-\dfrac{25}{80}\)
\(=\dfrac{30-\left(-72\right)-25}{80}\)
\(=\dfrac{77}{80}\)
TT
ai giúp mình bài 16 với mình cảm ơn nhìu
0
MD
0
TE
1
Bài 16:
1: \(\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}-2}-3\cdot\frac{\sqrt{x}-1}{x-5\sqrt{x}+6}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)-3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-4-\left(x-2\sqrt{x}-3\right)-3\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}-1-x+2\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{-1}{\sqrt{x}-3}\)
2: \(\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{\sqrt{x}-1}+\frac{x-2}{x-3\sqrt{x}+2}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)-\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4\sqrt{x}+3-\left(2x-5\sqrt{x}+2\right)+x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{2x-4\sqrt{x}+1-2x+5\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\frac{1}{\sqrt{x}-2}\)
3: \(\frac{3\sqrt{x}+2}{2\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+4}-\frac{x-6\sqrt{x}+5}{2x+7\sqrt{x}-4}\)
\(=\frac{\left(3\sqrt{x}+2\right)\left(\sqrt{x}+4\right)+\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)-x+6\sqrt{x}-5}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)
\(=\frac{3x+14\sqrt{x}+8+2x-3\sqrt{x}+1-x+6\sqrt{x}-5}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}=\frac{4x+17\sqrt{x}+4}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)
\(=\frac{\left(4\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}=\frac{4\sqrt{x}+1}{2\sqrt{x}-1}\)
4: \(\frac{\sqrt{x}+4}{1-7\sqrt{x}}+\frac{\sqrt{x}-2}{\sqrt{x}+1}+\frac{24\sqrt{x}}{7x+6\sqrt{x}-1}\)
\(=\frac{-\left(\sqrt{x}+4\right)\left(\sqrt{x}+1\right)+\left(\sqrt{x}-2\right)\left(7\sqrt{x}-1\right)+24\sqrt{x}}{\left(7\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-x-5\sqrt{x}-4+7x-15\sqrt{x}+2+24\sqrt{x}}{\left(7\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{6x+4\sqrt{x}-2}{\left(7\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(6\sqrt{x}-2\right)}{\left(7\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{6\sqrt{x}-2}{7\sqrt{x}-1}\)
5: \(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\)
\(=\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{9-x+x-9+\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}-2}{\sqrt{x}+3}\)
6: \(\frac{3-\sqrt{x}}{5\sqrt{x}+7}+\frac{3\sqrt{x}+4}{3\sqrt{x}-2}-\frac{42\sqrt{x}+34}{15x+11\sqrt{x}-14}\)
\(=\frac{\left(-\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)+\left(3\sqrt{x}+4\right)\left(5\sqrt{x}+7\right)-42\sqrt{x}-34}{\left(5\sqrt{x}+7\right)\left(3\sqrt{x}-2\right)}\)
\(=\frac{-3x+11\sqrt{x}-6+15x+41\sqrt{x}+28-42\sqrt{x}-34}{\left(5\sqrt{x}+7\right)\left(3\sqrt{x}-2\right)}=\frac{12x+10\sqrt{x}-12}{\left(5\sqrt{x}+7\right)\left(3\sqrt{x}-2\right)}\)
\(=\frac{12x-8\sqrt{x}+18\sqrt{x}-12}{\left(5\sqrt{x}+7\right)\left(3\sqrt{x}-2\right)}=\frac{\left(3\sqrt{x}-2\right)\left(4\sqrt{x}+6\right)}{\left(5\sqrt{x}+7\right)\left(3\sqrt{x}-2\right)}=\frac{4\sqrt{x}+6}{5\sqrt{x}+7}\)
7: \(\frac{\sqrt{x}+2}{-\sqrt{x}+2}+\frac{3\sqrt{x}-4}{2\sqrt{x}-3}+\frac{-7\sqrt{x}+10}{-2x+7\sqrt{x}-6}\)
\(=\frac{\left(-\sqrt{x}-2\right)\left(2\sqrt{x}-3\right)+\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)+7\sqrt{x}-10}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{-2x-\sqrt{x}+6+3x-10\sqrt{x}+8+7\sqrt{x}-10}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{x-4\sqrt{x}+4}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)^2}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}-2}{2\sqrt{x}-3}\)
8: \(-\frac{7\sqrt{x}+7}{5\sqrt{x}-1}+\frac{2\sqrt{x}-2}{\sqrt{x}+2}+\frac{39\sqrt{x}+12}{5x+9\sqrt{x}-2}\)
\(=\frac{-\left(7\sqrt{x}+7\right)\left(\sqrt{x}+2\right)+\left(2\sqrt{x}-2\right)\left(5\sqrt{x}-1\right)+39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{-7x-21\sqrt{x}-14+10x-12\sqrt{x}+2+39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{3x+6\sqrt{x}}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{3\sqrt{x}}{5\sqrt{x}-1}\)
9: \(\frac{-5\sqrt{x}+4}{3\sqrt{x}-2}+\frac{6\sqrt{x}+4}{2\sqrt{x}+3}+\frac{29\sqrt{x}-28}{3\left(6x+5\sqrt{x}-6\right)}\)
\(=\frac{\left(-5\sqrt{x}+4\right)\left(2\sqrt{x}+3\right)+\left(6\sqrt{x}+4\right)\left(3\sqrt{x}-2\right)}{\left(3\sqrt{x}-2\right)\left(2\sqrt{x}+3\right)}+\frac{29\sqrt{x}-28}{3\left(3\sqrt{x}-2\right)\left(2\sqrt{x}+3\right)}\)
\(=\frac{-10x-7\sqrt{x}+12+18x-8}{\left(3\sqrt{x}-2\right)\left(2\sqrt{x}+3\right)}+\frac{29\sqrt{x}-28}{3\left(3\sqrt{x}-2\right)\left(2\sqrt{x}+3\right)}\)
\(=\frac{8x-7\sqrt{x}+4}{\left(3\sqrt{x}-2\right)\left(2\sqrt{x}+3\right)}+\frac{29\sqrt{x}-28}{3\left(3\sqrt{x}-2\right)\left(2\sqrt{x}+3\right)}\)
\(=\frac{24x-21\sqrt{x}+12}{3\left(3\sqrt{x}-2\right)\left(2\sqrt{x}+3\right)}+\frac{29\sqrt{x}-28}{3\left(3\sqrt{x}-2\right)\left(2\sqrt{x}+3\right)}=\frac{24x+8\sqrt{x}-16}{3\left(3\sqrt{x}-2\right)\left(2\sqrt{x}+3\right)}\)
\(=\frac{8\left(3x+\sqrt{x}-2\right)}{3\left(3\sqrt{x}-2\right)\left(2\sqrt{x}+3\right)}=\frac{8\left(3\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{3\left(3\sqrt{x}-2\right)\left(2\sqrt{x}+3\right)}=\frac{8\left(\sqrt{x}+1\right)}{3\left(2\sqrt{x}+3\right)}\)
10: \(\frac{2\sqrt{x}-4}{3\sqrt{x}-4}-\frac{2\sqrt{x}+4}{\sqrt{x}-2}+\frac{x+13\sqrt{x}-20}{3x-10\sqrt{x}+8}\)
\(=\frac{\left(2\sqrt{x}-4\right)\left(\sqrt{x}-2\right)-\left(2\sqrt{x}+4\right)\left(3\sqrt{x}-4\right)+x+13\sqrt{x}-20}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{2x-8\sqrt{x}+8-\left(6x+4\sqrt{x}-16\right)+x+13\sqrt{x}-20}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{3x+5\sqrt{x}-12-6x-4\sqrt{x}+16}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}=\frac{-3x+\sqrt{x}+4}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{-3x+4\sqrt{x}-3\sqrt{x}+4}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}=\frac{\left(3\sqrt{x}-4\right)\left(-\sqrt{x}-1\right)}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}=\frac{-\sqrt{x}-1}{\sqrt{x}-2}\)