a, \(=\dfrac{5}{2}+\dfrac{15}{4}=\dfrac{25}{4}\)

b, \(=3-\dfrac{7}{5}=\dfrac{8}{5}\)

c, \(=\dfrac{23}{5}.\dfrac{100}{46}=\dfrac{20}{2}=10\)

d, \(=\dfrac{72}{17}:9=\dfrac{8}{17}\)

a, \(=\dfrac{13}{5}+\dfrac{5}{4}.\dfrac{8}{3}=\dfrac{13}{5}+\dfrac{10}{3}=\dfrac{89}{15}\)

b, \(=\dfrac{36}{7}-\dfrac{14}{5}:\dfrac{6}{5}=\dfrac{36}{7}-\dfrac{7}{3}=\dfrac{59}{21}\)

c, \(=\dfrac{17}{4}-\dfrac{21}{8}.\dfrac{16}{39}=\dfrac{17}{4}-\dfrac{42}{39}=\dfrac{165}{52}\)

:> em mà 

Ai bình chọn cho mình. Mình chuyển khoản 1 coin tiền thưởng luôn nhé ;)) 

A/c cho em hỏi với. Tự bình chọn cho mình được ko ạ ? :(( 

b, \(\left|x-4\right|=4-x\)đk x =<4 

=> x - 4 = 4 - x <=> x = 4 (tm) 

c, \(\left|x+1\right|+2\left|x+1\right|=3x\Leftrightarrow3\left|x+1\right|=3x\Leftrightarrow\left|x+1\right|=x\)

đk x >= 0

TH1 : x + 1 = -x <=> 2x = -1 <=> x = -1/2 (ktm)

TH2 : x + 1 = x <=> 0 = -1 (vô lí)

 d, \(\left|x-1\right|=3x-2\)đk x >= 2/3

TH1 : x - 1 = 3x - 2 <=> x = 1/2 (ktm) 

TH2 : x - 1 = 2 - 3x <=> 4x = 3 <=> x = 3/4 (tm) 

đk x > = 0 ; x khác 9 ; 64 

a, Thay vào ta được 

\(P=\left(\dfrac{4}{4+3}-\dfrac{2\left(4+12\right)}{16-9}\right).\dfrac{4+5}{4-8}=\left(\dfrac{4}{7}-\dfrac{32}{17}\right).\dfrac{9}{-4}\)

\(=-\dfrac{9}{7}+\dfrac{72}{17}=\dfrac{351}{119}\)

b, \(P=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-2\sqrt{x}-24}{x-9}\right).\dfrac{\sqrt{x}+5}{\sqrt{x}-8}\)

\(=\dfrac{x-5\sqrt{x}-24}{x-9}.\dfrac{\sqrt{x}+5}{\sqrt{x}-8}=\dfrac{\sqrt{x}+5}{\sqrt{x}-3}\)

c, Ta có \(P-1=\dfrac{\sqrt{x}+5}{\sqrt{x}-3}-1=\dfrac{8}{\sqrt{x}-3}\le0\Rightarrow\sqrt{x}-3< 0\Leftrightarrow x< 9\)

Kết hợp đk vậy 0 =< x < 9 

a, đk m;n;x;y >= 0 

\(\sqrt{mx}-\sqrt{ny}+\sqrt{nx}-\sqrt{my}=\sqrt{m}\left(\sqrt{x}-\sqrt{y}\right)-\sqrt{n}\left(\sqrt{y}-\sqrt{x}\right)=\left(\sqrt{m}+\sqrt{n}\right)\left(\sqrt{x}-\sqrt{y}\right)\)

b, \(6x\left(x^2-2\right)=6x\left(\sqrt{x}-2\right)\left(\sqrt{x}+\sqrt{2}\right)\)

c, \(C=\left|x-1\right|+\left|3-x\right|\ge\left|x-1+3-x\right|=2\)

Dấu ''='' xảy ra khi 1 =< x =< 3 

d, \(D=\left|x+1\right|+\left|4-x\right|=\left|x+1+4-x\right|=5\)

Dấu ''='' xảy ra khi -1 =< x =< 4 

e, \(E=\left|x-10\right|+\left|x-3\right|+\left|x-5\right|\ge7+\left|x-5\right|\)

Với x >= 10 => x - 5 >= 5 <=> |x-5| >= 5 

<=> |x-5| + 7 >= 12 

Dấu ''='' xảy ra khi x = 10 

1, \(4y^2\left(x^2+y^2\right)-5xy\left(x^2+y^2\right)=y\left(4y-5x\right)\left(x^2+y^2\right)\)

2,\(\left(x^2-x+2\right)\left(x-1\right)-x^2\left(x-1\right)^2-\left(2x+1\right)\left(1-x\right)^3\)

\(=\left(x-1\right)\left(x^2-x+2\right)-\left(x^3-x^2\right)\left(x-1\right)+\left(2x+1\right)\left(1-x\right)^2\left(x-1\right)\)

\(=\left(x-1\right)\left[x^2-x+2-x^3+x^2+\left(2x+1\right)\left(x^2-2x+1\right)\right]\)

\(=\left(x-1\right)\left(-x^3-x+2+2x^3-4x^2+2x+x^2-2x+1\right)=\left(x-1\right)\left(x^3-3x^2-x+3\right)\)

\(=\left(x-1\right)\left[x^2\left(x-3\right)-\left(x-3\right)\right]=\left(x-1\right)^2\left(x+1\right)\left(x-3\right)\)