a: Ta có: \(ax-x+1=a^2\)

\(\Leftrightarrow x\left(a-1\right)=a^2-1\)

hay x=a+1

P(x)=\(ax^2+bx+c\) (1)(a\(\ne0\) )

Ta có : 

\(\dfrac{a}{1}=\dfrac{b}{2}=\dfrac{c}{3}\)\(\Rightarrow\left\{{}\begin{matrix}b=2a\\c=3a\end{matrix}\right.\)(2)

Thay(2) vào (1)\(\Rightarrow P\left(x\right)=ax^2+2ax+3a\)

\(\Rightarrow\dfrac{P\left(-2\right)-3P\left(-1\right)}{a}=\dfrac{4a-4a+3a-3\left(a-2a+3a\right)}{a}\)=\(\dfrac{3a-3a+6a-9a}{a}=\dfrac{-3a}{a}=-3\)

Câu b bạn sửa lại đề

\(a,VT=\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=VP\\ b,VT=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\ =\sqrt{a}-\sqrt{b}+\sqrt{a}+\sqrt{b}=2\sqrt{a}=VP\)

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

Với \(c=0\Rightarrow f\left(x\right)=0\) có nghiệm \(x=0\) (loại)

TH1: \(a;c\) trái dấu 

Xét pt \(f\left(x\right)=0\Leftrightarrow a\left(ax^2+bx+c\right)^2+b\left(ax^2+bx+c\right)+c=0\)

Đặt \(ax^2+bx+c=t\) \(\Rightarrow at^2+bt+c=0\) (1)

Do a; c trái dấu \(\Leftrightarrow\) (1) luôn có 2 nghiệm trái dấu.

Không mất tính tổng quát, giả sử \(t_1< 0< t_2\)

\(\Rightarrow\left[{}\begin{matrix}ax^2+bx+c=t_1\\ax^2+bx+c=t_2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}ax^2+bx+c-t_1=0\left(2\right)\\ax^2+bx+c-t_2=0\left(3\right)\end{matrix}\right.\)

Mà a; c trái dấu nên:

- Nếu \(a>0\Rightarrow c< 0\Rightarrow c-t_2< 0\Rightarrow a\left(c-t_2\right)< 0\)

\(\Rightarrow\) (3) có nghiệm hay \(f\left(x\right)=0\) có nghiệm (loại)

- Nếu \(a< 0\Rightarrow c>0\Rightarrow c-t_1>0\Rightarrow a\left(c-t_1\right)< 0\)

\(\Rightarrow\left(2\right)\) có nghiệm hay \(f\left(x\right)=0\) có nghiệm (loại)

Vậy đa thức \(f\left(x\right)\) luôn có nghiệm khi a; c trái dấu

\(\Rightarrow\)Để \(f\left(x\right)=0\) vô nghiệm thì điều kiện cần là \(a;c\) cùng dấu \(\Leftrightarrow ac>0\)

Khi đó xét \(g\left(x\right)=0\) có \(a.\left(-c\right)< 0\Rightarrow g\left(x\right)=0\) luôn có 2 nghiệm trái dấu (đpcm)

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

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

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

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

\(A=-1\Leftrightarrow1-x=-1\Rightarrow x=2\)

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

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

=1-x

b) Để A=-1 thì 1-x=-1

hay x=2

\(a,A=\dfrac{2\sqrt{x}-2-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ A=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{1}{\sqrt{x}-1}\\ b,A< 0\Leftrightarrow\dfrac{1}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\left(1>0\right)\\ \Leftrightarrow x< 1\\ c,A\in Z\Leftrightarrow1⋮\sqrt{x}-1\\ \Leftrightarrow\sqrt{x}-1\inƯ\left(1\right)\left\{-1;1\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{0;2\right\}\\ \Leftrightarrow x\in\left\{0;4\right\}\)

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

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

b) \(A=\dfrac{1}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\)

Kết hợp đk: 

\(\Rightarrow0\le x< 1\)

c) \(A=\dfrac{1}{\sqrt{x}-1}\in Z\)

\(\Rightarrow\sqrt{x}-1\inƯ\left(1\right)=\left\{-1;1\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{0;2\right\}\)

\(\Rightarrow x\in\left\{0;4\right\}\)

Bài 1:

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)

Khi đó: \(\left\{\begin{matrix} \frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b(2k+5)}{b(3k-4)}=\frac{2k+5}{3k-4}\\ \frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d(2k+5)}{d(3k-4)}=\frac{2k+5}{3k-4}\end{matrix}\right.\)

\(\Rightarrow \frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)

Ta có đpcm.

Bài 2:

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)

Khi đó: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)

\(\frac{a^2+b^2}{c^2+d^2}=\frac{(bk)^2+b^2}{(dk)^2+d^2}=\frac{b^2(k^2+1)}{d^2(k^2+1)}=\frac{b^2}{d^2}\)

Do đó: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}(=\frac{b^2}{d^2})\) . Ta có đpcm.

Bài 3:

a) Sửa điều kiện: \(\frac{a}{b}=\frac{c}{d}\neq -1\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)

Theo đkđb thì \(k\neq -1\) nên \(k^3+1\neq 0\); \(k+1\neq 0\)

Ta có: \(\frac{a^3+b^3}{c^3+d^3}=\frac{(bk)^3+b^3}{(dk)^3+d^3}=\frac{b^3(k^3+1)}{d^3(k^3+1)}=\frac{b^3}{d^3}\)

\(\frac{(a+b)^3}{(c+d)^3}=\frac{(bk+b)^3}{(dk+d)^3}=\frac{b^3(k+1)^3}{d^3(k+1)^3}=\frac{b^3}{d^3}\)

\(\Rightarrow \frac{a^3+b^3}{c^3+d^3}=\frac{(a+b)^3}{(c+d)^3}\) (đpcm)

b)

Đặt \(\frac{a}{b}=k; \frac{c}{d}=t\Rightarrow a=bk; c=dt\)

Ta cần cm \(k=t\)

Khi đó:

\(\frac{2a+13b}{3a-7b}=\frac{2bk+13b}{3bk-7b}=\frac{b(2k+13)}{b(3k-7)}=\frac{2k+13}{3k-7}\)

\(\frac{2c+13d}{3c-7d}=\frac{2dt+13d}{3dt-7d}=\frac{d(2t+13)}{d(3t-7)}=\frac{2t+13}{3t-7}\)

Vì \(\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}\Rightarrow \frac{2k+13}{3k-7}=\frac{2t+13}{3t-7}\)

\(\Rightarrow (2k+13)(3t-7)=(2t+13)(3k-7)\)

\(-14k+39t=-14t+39k\Rightarrow k=t\)

Ta có đpcm.