giúp mình với

em ăn nhiều bánh nhất

em ăn \(\dfrac{1}{2}=\dfrac{3}{6}\) cái bánh

chị ăn \(\dfrac{2}{3}=\dfrac{4}{6}\) cái bánh

⇒chị ăn nhiều bánh hơn

(cách mình trình bày chắc chắn là không đúng nên bạn trình bày lại theo cách cô chỉ nha!)

a) Để phương trình có nghiệm thì: m²≠0 =>m≠0

b) Vì phương trình có nghiệm bằng -2m

=>-(m+2)2m-m²=0 ⇔-2m²-4m-m²=0 ⇔-3m²-4m=0 ⇔-m(3m+4)=0 

⇔m=0 hay m=\(\dfrac{-4}{3}\)mà m phải khác 0 nên m=\(\dfrac{-4}{3}\).

c) -(m+2)x-m²=0 ⇔x=\(\dfrac{m^2}{m+2}\)>0 ⇔m+2>0 ⇔m>-2.

d)  -(m+2)x-m²=0 ⇔x=\(\dfrac{m^2}{m+2}\).

Để x nguyên thì m² ⋮ m+2.

⇔ m²-4+4 ⋮ m+2

⇔ 4 ⋮ m+2

⇔ m∈{-1;-3;0;-4;2;-6} mà m khác 0 nên m∈{-1;-3;-4;2;-6}

\(\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\\ =\dfrac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\dfrac{2+\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\\ =\dfrac{2-\sqrt{3}+2+\sqrt{3}}{2^2-\left(\sqrt{3}\right)^2}\\ =\dfrac{2+2}{4-3}\\ =4\)

Ta có: \(\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\)

\(=2-\sqrt{3}+2+\sqrt{3}\)

=4

(3x)^2=3^2.x^2=9x^2

Ta có:

\(3x-3=3\left(x-1\right)\)

\(4-4x=-4\left(x-1\right)\)

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

\(\Rightarrow\) MTC là \(3.\left(-4\right).\left(x-1\right)\left(x+1\right)=-12\left(x-1\right)\left(x+1\right)\)

Do đó:

\(\dfrac{11x}{3x-3}=\dfrac{11x}{3\left(x-1\right)}=\dfrac{11x.\left(-4\right).\left(x+1\right)}{3\left(x-1\right).\left(-4\right)\left(x+1\right)}=\dfrac{-44x\left(x+1\right)}{-12\left(x-1\right)\left(x+1\right)}\)

\(\dfrac{5}{4-4x}=\dfrac{5}{-4\left(x-1\right)}=\dfrac{5.3\left(x+1\right)}{-4\left(x-1\right).3\left(x+1\right)}=\dfrac{15\left(x+1\right)}{-12\left(x-1\right)\left(x+1\right)}\)

\(\dfrac{2x}{x^2-1}=\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=\dfrac{2x.\left(-12\right)}{-12\left(x-1\right)\left(x+1\right)}=\dfrac{-24x}{-12\left(x-1\right)\left(x+1\right)}\)

\(\dfrac{2}{2x+1}-1\ge0\Leftrightarrow\dfrac{2}{2x+1}-\dfrac{2x+1}{2x+1}\ge0\)

\(\Leftrightarrow\dfrac{2-\left(2x+1\right)}{2x+1}\ge0\)

\(\Leftrightarrow\dfrac{1-2x}{2x+1}\ge0\)

\(\Rightarrow-\dfrac{1}{2}< x\le\dfrac{1}{2}\)

b) Ta có: \(\sqrt{150}-\sqrt{1.6}\cdot\sqrt{60}+4.5\cdot\sqrt{2\dfrac{2}{3}}-\sqrt{6}\)

\(=5\sqrt{6}-4\sqrt{6}-\sqrt{6}+\dfrac{9}{2}\cdot\sqrt{\dfrac{8}{3}}\)

\(=\dfrac{9}{2}\cdot\dfrac{2\sqrt{2}}{\sqrt{3}}\)

\(=3\sqrt{6}\)

\(\sqrt{150}+\sqrt{1,6}.\sqrt{60}+4.5\sqrt{2\dfrac{2}{3}}-\sqrt{6}\\ =5\sqrt{6}+4\sqrt{6}+3\sqrt{6}-\sqrt{6}\\ =11\sqrt{6}\)

a: ĐKXĐ: \(x\le0\)

b: ĐKXĐ: \(x\le2\)

\(a,ĐK:-3x\ge0\Leftrightarrow x\le0\left(-3< 0\right)\\ b,ĐK:4-2x\ge0\Leftrightarrow-2x\ge-4\Leftrightarrow x\le2\\ c,ĐK:\dfrac{1}{2x-5}\ge0\Leftrightarrow2x-5>0\left(1>0;2x-5\ne0\right)\\ \Leftrightarrow x>\dfrac{5}{2}\\ d,ĐK:\dfrac{4x+7}{-3}\ge0\Leftrightarrow4x+7\le0\left(-3< 0\right)\Leftrightarrow x\le-\dfrac{7}{4}\)