\(a,=\dfrac{x^3-\left(x-1\right)\left(x^2+x+1\right)}{1-x}=\dfrac{x^3-x^3+1}{1-x}=\dfrac{1}{1-x}\\ b,=\dfrac{2x+x^2+3x+2+2-x}{\left(x+2\right)^2}=\dfrac{\left(x+2\right)^2}{\left(x+2\right)^2}=1\)

mình có đáp án rồi ạ.

Câu 1:

\(\left(x-2\right)\left(x^2+2x+4\right)+25x=x\left(x+5\right)\left(x-5\right)+8\)

\(\Leftrightarrow x^3-8+25x=x\left(x^2-25\right)+8\)

\(\Leftrightarrow x^3-8+25x=x^3-25x+8\)

\(\Leftrightarrow x^3-8+25x-x^3+25x-8=0\)

\(\Leftrightarrow50x-16=0\)

\(\Leftrightarrow50x=16\)

\(\Leftrightarrow x=\dfrac{8}{25}\)

Câu 2 :

\(\dfrac{x+5}{4}+\dfrac{3+2x}{3}=\dfrac{6x-1}{3}-\dfrac{1-2x}{12}\)

<=> \(\dfrac{3\left(x+5\right)}{12}+\dfrac{4\left(3+2x\right)}{12}=\dfrac{4\left(6x-1\right)}{12}-\dfrac{1-2x}{12}\)

<=>\(\dfrac{3x+15+12+8x}{12}=\dfrac{24x-4-1+2x}{12}\)

<=> 3x + 15 + 12 + 8x = 24x - 4 - 1 +2x

<=> 11x+27 = 26x -5

<=> ( 26x - 5 ) - ( 11x + 27 ) = 0

<=> 15x - 32 = 0

<=> 15x = 32

<=> x = \(\dfrac{32}{15}\)

Câu 3:

x - 4/3 - 3x - 1/12 = 3x + 1/4 + 9x - 2/8

<=> 4x - 16 - 3x + 1/12 = 6x + 2 + 9x - 2/8

<=> x - 15/12 = 15x/8

<=> 8x - 120 = 180x

<=> 120 = -172x <=> x = -172/120 = -43/30

1) \(\sqrt{4+x}=2-x\) (ĐK: \(x\ge-4\))

\(\Leftrightarrow4+x=\left(2-x\right)^2\)

\(\Leftrightarrow4+x=4-4x+x^2\)

\(\Leftrightarrow x^2-4x-x+4-4=0\)

\(\Leftrightarrow x^2-5x=0\)

\(\Leftrightarrow x\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

Vậy: \(S=\left\{0;5\right\}\)

2) 

a) ĐKXĐ: \(a>0,a\ne1\)

\(A=\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a}\)

\(A=\left[\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right]\cdot\dfrac{a}{\sqrt{a}+1}\)

\(A=\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\cdot\dfrac{a}{\sqrt{a}+1}\)

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

\(A=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\cdot\dfrac{\sqrt{a}\cdot\sqrt{a}}{\sqrt{a}+1}\)

\(A=\sqrt{a}\left(\sqrt{a}-1\right)\)

\(A=a-\sqrt{a}\)

b) Ta có:

\(A=a-\sqrt{a}\)

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

\(A=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)

Mà: \(\left(\sqrt{a}-\dfrac{1}{2}\right)^2\ge0\) nên \(A=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra khi:

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

\(\Leftrightarrow a=\dfrac{1}{4}\)

Vậy: \(A_{min}=-\dfrac{1}{4}\)khi \(a=\dfrac{1}{4}\)

a: ĐKXĐ: x<>0; x<>-1

PT =>x+1-2x=3

=>1-x=3

=>x=-2(nhận)

b: Sửa đề: \(\dfrac{1}{2x-3}-\dfrac{3}{x\left(2x-3\right)}=\dfrac{5}{x}\)

=>x-3=5(2x-3)

=>10x-15=x-3

=>9x=12

=>x=4/3(nhận)

c: ĐKXĐ: x<>0; x<>2

PT =>x(x+2)-x+2=2

=>x^2+2x-x=0

=>x(x+1)=0

=>x=-1

2.

b, \(-4< \dfrac{2x^2+mx-4}{-x^2+x-1}< 6\)

\(\Leftrightarrow\left\{{}\begin{matrix}-4< \dfrac{2x^2+mx-4}{-x^2+x-1}\left(1\right)\\\dfrac{2x^2+mx-4}{-x^2+x-1}< 6\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow4\left(x^2-x+1\right)>2x^2+mx-4\)

\(\Leftrightarrow2x^2-\left(m+4\right)x+8>0\)

Yêu cầu bài toán thỏa mãn khi \(\Delta=m^2+8m-48< 0\Leftrightarrow-12< m< 4\)

\(\left(2\right)\Leftrightarrow-6\left(x^2-x+1\right)< 2x^2+mx-4\)

\(\Leftrightarrow8x^2+\left(m-6\right)x+2>0\)

Yêu cầu bài toán thỏa mãn khi \(\Delta=m^2-12m-28< 0\Leftrightarrow-2< x< 14\)

Vậy \(m\in\left(-2;4\right)\)

2.

a, Yêu cầu bài toán thỏa mãn khi phương trình \(\left(m-4\right)x^2+\left(1+m\right)x+2m-1>0\) có nghiệm đúng với mọi x

\(\Leftrightarrow\left\{{}\begin{matrix}m-4>0\\\Delta=m^2+2m+1-4\left(m-4\right)\left(2m-1\right)< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m>4\\\left[{}\begin{matrix}m< \dfrac{3}{7}\\m>5\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow m>5\)

1.

Yêu cầu bài toán thỏa mãn khi:

\(\left\{{}\begin{matrix}m-4< 0\\\Delta=-7m^2+38m-15< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< 4\\\left[{}\begin{matrix}m>5\\m< \dfrac{3}{7}\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow m< \dfrac{3}{7}\)

Hiển nhiên là cách đầu sai rồi em

Khi đến \(\lim x^2\left(1-1\right)=+\infty.0\) là 1 dạng vô định khác, đâu thể kết luận nó bằng 0 được

a: \(A=x^3+3x^2-5x-15+x^2-x^3+4x-4x^2\)

\(=-x-15\)

\(=-\left(-1\right)-15=1-15=-14\)

\(=\dfrac{5}{6}+\dfrac{1}{4}=\dfrac{13}{12}\)

\(\dfrac{13}{12}\)

5/6+ 1/4 = 13/12