\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\Leftrightarrow\frac{ab+bc+ac}{abc}=1\Leftrightarrow ab+bc+ac=abc\)

kết hợp gt: a+b+c=1

\(\Rightarrow abc-ab-ac-bc+a+b+c-1=0\Leftrightarrow\left(a-1\right)\left(b-1\right)\left(c-1\right)=0\left(đpcm\right)\)

\(\Leftrightarrow x^2+\left(2\sqrt{2}-3\right)x+4+3\sqrt{2}=0\)

\(\Delta=\left(2\sqrt{2}-3\right)^2-4\left(4+3\sqrt{2}\right)=1-24\sqrt{2}< 0\)

=> Pt vô nghiệm

Câu 1 :

\(a,2\left(\frac{3}{4}-5x\right)=\frac{4}{5}-3x\)

\(\Rightarrow\frac{3}{2}-10x=\frac{4}{5}-3x\)

\(\Rightarrow7x=\frac{3}{2}-\frac{4}{5}\)

\(\Rightarrow7x=\frac{7}{10}\)\(\Leftrightarrow x=0,1\)

\(b,\frac{3}{2}-4\left(\frac{1}{4}-x\right)=\frac{2}{3}-7x\)

\(\Rightarrow\frac{3}{2}-1+4x=\frac{2}{3}-7x\)

\(\Rightarrow11x=\frac{2}{3}+1-\frac{3}{2}\)

\(\Rightarrow11x=\frac{4+6-9}{6}-\frac{1}{6}\)

\(\Rightarrow x=\frac{1}{66}\)

Câu 2 :

\(a,\frac{2}{x-1}< 0\)

Vì \(2>0\Rightarrow\)để \(\frac{2}{x-1}< 0\)thì \(x-1< 0\Leftrightarrow x< 1\)

\(b,\frac{-5}{x-1}< 0\)

Vì \(-5< 0\)\(\Rightarrow\)để \(\frac{-5}{x-1}< 0\)thì \(x-1>0\Rightarrow x>1\)

\(c,\frac{7}{x-6}>0\)

Vì \(7>0\Rightarrow\)để \(\frac{7}{x-6}>0\)thì \(x-6>0\Rightarrow x>6\)

\(|x+1|=|3x-2|\)

TH1 : \(x+1=3x-2\)

\(\Rightarrow2x=3\)

\(\Rightarrow x=\frac{3}{2}\)

TH2 : \(x+1=-3x+2\)

\(\Rightarrow4x=1\)

\(\Rightarrow x=\frac{1}{4}\)

Vậy \(x\in\left\{\frac{3}{2};\frac{1}{4}\right\}\)

\(|x+1|=|3x-2|\)

\(Th1:x+1=3x-2\)

\(\Rightarrow2x=3\Rightarrow x=\frac{3}{2}\)

\(Th2:x+1=2-3x\)

\(\Rightarrow4x=1\Rightarrow x=\frac{1}{4}\)

\(\left(\frac{1}{x-\sqrt{x}}-\frac{\sqrt{x}}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\sqrt{x}}\)

\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\frac{\sqrt{x}+1}{\sqrt{x}}\)

\(=\frac{1-x}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}}{\sqrt{x}+1}\)

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

TL:

= - 1

-HT-

!!!!!

Mik làm luôn, ko vt lại đề

B = 1/1 - 1/3 + 1/3 - 1/5 +...+1/2017 - 1/2019

B = 1 - 1/2019

B = 2018/2019

TL:

\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2019}\) 

\(B=1-\frac{1}{2019}\) 

\(B=\frac{2018}{2019}\) 

Vậy \(B=\frac{2018}{2019}\)

a) Ta có:

C = 5/18 + 8/19 - 7/21 + (-10/36 + 11/19 + 1/3) - 5/8

C = 5/18 + 8/19 - 1/3 - 5/18 + 11/19 + 1/3 - 5/8

C = (5/18 - 5/18) + (8/19 + 11/19) - (1/3  - 1/3) - 5/8

C = 1 - 5/8

c = 3/8

b) F = 15/14 - (17/23 - 80/87 + 5/4) + (17/23 - 15/14 + 1/4)

F = 15/14 - 17/23 + 80/87 - 5/4 + 17/23 - 15/14 + 1/4

F = (15/14 - 15/14) - (17/23 - 17/23) + 80/87 - (5/4 - 1/4)

F = 80/87 - 1

F = -7/87

c) G = 1/25 - 4/27 + (-23/27 + -1/25 - 5/43) + 5/43 - 4/7

G = 1/25 - 4/27 - 23/27 - 1/25 - 5/43 + 5/43 - 4/7

G = (1/25 - 1/25) - (4/27 + 23/27) - (5/43 - 5/43) - 4/7

G = -1 - 4/7 = -11/7

trần đắc lợi lần sau nhớ gõ latex nha bạn, như này người làm dễ bị sai đề lắm

\(a\sqrt{b-1}+b\sqrt{a-1}\)

Áp dụng AM-GM :

\(a\sqrt{b-1}+b\sqrt{a-1}\)

\(=a\sqrt{1\cdot\left(b-1\right)}+b\sqrt{1\cdot\left(a-1\right)}\le a\cdot\frac{1+b-1}{2}+b\cdot\frac{1+a-1}{2}\)

\(=\frac{ab}{2}+\frac{ab}{2}=ab\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=2\)

cảm ơn nha