[2x-2=0=>x=1

x-1=0=>x=1

x+1=0=>x=-1

5=0=>x=5

a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)

=1-2/4=1/2

b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)

c: x-y=0 nên x=y

\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)

=2019

Bài 1: ĐKXĐ:`x + 3 ne 0` và `x^2+ x-6 ne 0 ; 2-x ne 0`

`<=> x ne -3 ; (x-2)(x+3) ne 0 ; x ne2`

`<=>x ne -3 ; x ne 2`

b) Với `x ne - 3 ; x ne 2` ta có:

`P= (x+2)/(x+3)  - 5/(x^2 +x -6) + 1/(2-x)`

`P = (x+2)/(x+3) - 5/[(x-2)(x+3)] + 1/(2-x)`

`= [(x+2)(x-2)]/[(x-2)(x+3)] - 5/[(x-2)(x+3)] - (x+3)/[(x-2)(x+3)]`

`= (x^2 -4)/[(x-2)(x+3)] - 5/[(x-2)(x+3)] - (x+3)/[(x-2)(x+3)]`

`=(x^2 - 4 - 5 - x-3)/[(x-2)(x+3)]`

`= (x^2 - x-12)/[(x-2)(x+3)]`

`= [(x-4)(x+3)]/[(x-2)(x+3)]`

`= (x-4)/(x-2)`

Vậy `P= (x-4)/(x-2)` với `x ne -3 ; x ne 2`

c) Để `P = -3/4`

`=> (x-4)/(x-2) = -3/4`

`=> 4(x-4) = -3(x-2)`

`<=>4x -16 = -3x + 6`

`<=> 4x + 3x = 6 + 16`

`<=> 7x = 22`

`<=> x= 22/7` (thỏa mãn ĐKXĐ)

Vậy `x = 22/7` thì `P = -3/4`

d) Ta có: `P= (x-4)/(x-2)`

`P= (x-2-2)/(x-2)`

`P= 1 - 2/(x-2)`

Để P nguyên thì `2/(x-2)` nguyên

`=> 2 vdots x-2`

`=> x -2 in Ư(2) ={ 1 ;2 ;-1;-2}`

+) Với `x -2 =1 => x= 3` (thỏa mãn ĐKXĐ)

+) Với `x -2 =2 => x= 4`  (thỏa mãn ĐKXĐ)

+) Với `x -2 = -1=> x= 1` (thỏa mãn ĐKXĐ)

+) Với `x -2 = -2 => x= 0`(thỏa mãn ĐKXĐ)

Vậy `x in{ 3 ;4; 1; 0}` thì `P` nguyên

e) Từ `x^2 -9 =0`

`<=> (x-3)(x+3)=0`

`<=> x= 3` hoặc `x= -3`

+) Với `x=3` (thỏa mãn ĐKXĐ) thì:

`P  = (3-4)/(3-2)`

`P= -1/1`

`P=-1`

+) Với `x= -3` thì không thỏa mãn ĐKXĐ

Vậy với x= 3 thì `P= -1`

\(\dfrac{5}{8}:\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{5}{8}\times\dfrac{4}{3}-\dfrac{1}{6}=\dfrac{5}{6}-\dfrac{1}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)

\(\left(\dfrac{3}{14}+\dfrac{1}{2}\right)\times\dfrac{7}{5}=\left(\dfrac{3}{14}+\dfrac{7}{14}\right)\times\dfrac{7}{15}=\dfrac{10}{14}\times\dfrac{7}{15}=\dfrac{5}{7}\times\dfrac{7}{15}=\dfrac{5}{12}=\dfrac{1}{3}\)

`a)` Thay `x=2` vào `B` có: `B=[-10]/[2-4]=5`

`b)` Với `x ne -1;x ne -5` có:

`A=[(x+2)(x+1)-5x-1-(x+5)]/[(x+1)(x+5)]`

`A=[x^2+x+2x+2-5x-1-x-5]/[(x+1)(x+5)]`

`A=[x^2-3x-4]/[(x+1)(x+5)]`

`A=[(x+1)(x-4)]/[(x+1)(x+5)]`

`A=[x-4]/[x+5]`

`c)` Với `x ne -5; x ne -1; x ne 4` có:

`P=A.B=[x-4]/[x+5].[-10]/[x-4]`

           `=[-10]/[x+5]`

Để `P` nguyên `<=>[-10]/[x+5] in ZZ`

    `=>x+5 in Ư_{-10}`

Mà `Ư_{-10}={+-1;+-2;+-5;+-10}`

`=>x={-4;-6;-3;-7;0;-10;5;-15}` (t/m đk)

= 13/5 x 1/4 x 3/2

= 13 x 1 x 3/ 5 x 4 x 2

= 39/40

=13/5x1/4x3/2

=13x1x3/5x4x2

=39/40

= \(\dfrac{13}{5}\)x\(\dfrac{1}{4}\)x\(\dfrac{3}{2}\)
=  \(\dfrac{39}{40}\)

Ai lm đc câu nào thì giúp mk với , cảm ơn !!

\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\\ A_{min}=\dfrac{1}{9}\Leftrightarrow x=\dfrac{3}{5}\\ B=\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\le\dfrac{2009}{2008}\\ B_{max}=\dfrac{2009}{2008}\Leftrightarrow x=\dfrac{3}{5}\\ C=-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\le1\dfrac{2}{3}\\ C_{max}=1\dfrac{2}{3}\Leftrightarrow\dfrac{1}{3}x=-4\Leftrightarrow x=-12\)

a: \(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{5}\)

\(a.x=3-2\sqrt{2}\\ \Rightarrow\sqrt{x}=\sqrt{3-2\sqrt{2}}\\ =\sqrt{2-2\sqrt{2}+1}\\ =\sqrt{\left(\sqrt{2}-1\right)^2}\\ =\left|\sqrt{2}-1\right|\\ =\sqrt{2}-1\left(vì\sqrt{2}>1\right)\)

Thay \(\sqrt{x}=\sqrt{2}-1\) vào A ta được

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

\(b.B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}-\dfrac{10-5\sqrt{x}}{x-5\sqrt{x}+6}\\ B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{10-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{10-5\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ B=\dfrac{x-3\sqrt{x}-\sqrt{x}+3-x+4-10+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{1}{\sqrt{x}-2}\)

\(c,P=A:B\\ P=\dfrac{\sqrt{x}}{1+\sqrt{x}}:\dfrac{1}{\sqrt{x}-2}\\ P=\dfrac{x-2\sqrt{x}}{1+\sqrt{x}}\)

\(P=\dfrac{-\sqrt{x}\left(-\sqrt{x}+2\right)}{\sqrt{x}+1}\)

Có: \(\sqrt{x}\ge0\)

\(\Rightarrow\sqrt{x}+1\ge1\left(I\right)\)

Lại có: \(\sqrt{x}\ge0\)

\(\Rightarrow-\sqrt{x}\le0\\ \Rightarrow-\sqrt{x}+2\le2\)

mà \(-\sqrt{x}\le0\)

\(\Rightarrow-\sqrt{x}\left(-\sqrt{x}+2\right)\ge2\)

Kết hợp với \(\left(I\right)\) \(\Rightarrow\) \(P=\dfrac{-\sqrt{x}\left(-\sqrt{x}+2\right)}{\sqrt{x}+1}\ge2\)

Vậy gtnn của P = \(2\) khi \(x=10+4\sqrt{6}\)

a: Khi \(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\) thì 

\(A=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{1+\sqrt{\left(\sqrt{2}-1\right)^2}}=\dfrac{\sqrt{2}-1}{1+\sqrt{2}-1}=\dfrac{\sqrt{2}-1}{\sqrt{2}}=\dfrac{2-\sqrt{2}}{2}\)

b: \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}-\dfrac{10-5\sqrt{x}}{x-5\sqrt{x}+6}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-4\sqrt{x}+3-x+4+5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}-2}\)

5/2 - 1/4 + 5/3

= 10/4 - 1/4 + 5/3

= 9/4 + 5/3

= 27/12 + 20/12

= 47/12

11/2 : 1/4 x 5/3

= 11/2 x 4/1 x 5/3

= 44/2 x 5/3

= 220/6

= 110/3

14/5 x 2/3 + 5

= 28/15 + 5

= 28/15 + 75/15

= 103/15

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