e: \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{3}{y}=3\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-7}{y}=-2\\\dfrac{1}{x}-\dfrac{1}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7}{2}\\\dfrac{1}{x}=1+\dfrac{2}{7}=\dfrac{9}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7}{2}\\x=\dfrac{7}{9}\end{matrix}\right.\)

c) \(\left\{{}\begin{matrix}2\left(x-2\right)+3\left(1+y\right)=2\\3\left(x-2\right)-2\left(1+y\right)=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6\left(x-2\right)+9\left(1+y\right)=6\\6\left(x-2\right)-4\left(1+y\right)=-6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}13\left(1+y\right)=12\\2\left(x-2\right)+3\left(1+y\right)=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{21}{13}\\y=-\dfrac{1}{13}\end{matrix}\right.\)

d) \(\left\{{}\begin{matrix}\left(x-5\right)\left(y-2\right)=\left(x+2\right)\left(y-1\right)\\\left(x-4\right)\left(y+7\right)=\left(x-3\right)\left(y+4\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy-2x-5y+10=xy-x+2y-2\\xy+7x-4y-28=xy+4x-3y-12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-x-7y=-12\\3x-y=16\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-x-7y=-12\\21x-7y=112\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}22x=124\\3x-y=16\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{62}{11}\\y=\dfrac{10}{11}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\left(x-5\right)\left(y-2\right)=\left(x+2\right)\left(y-1\right)\\\left(x-4\right)\left(y+7\right)=\left(x-3\right)\left(y+4\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy-2x-5y+10=xy-x+2y-2\\xy+7x-4y-28=xy+4x-3y-12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+7y=12\\3x-y=16\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x+21y=36\\3x-y=16\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}22y=20\\x+7y=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{62}{11}\\y=\dfrac{10}{11}\end{matrix}\right.\)

4:

x+3y=4m+4 và 2x+y=3m+3

=>2x+6y=8m+8 và 2x+y=3m+3

=>5y=5m+5 và x+3y=4m+4

=>y=m+1 và x=4m+4-3m-3=m+1

x+y=4

=>m+1+m+1=4

=>2m+2=4

=>2m=2

=>m=1

3:

x+2y=3m+2 và 2x+y=3m+2

=>2x+4y=6m+4 và 2x+y=3m+2

=>3y=3m+2 và x+2y=3m+2

=>y=m+2/3 và x=3m+2-2m-4/3=m+2/3

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\2\left(x+y\right)+4\left(x-y\right)=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=6\\2\left(x+y\right)+24=10\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=6\\x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{13}{2}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{x+y}{5}=\dfrac{x-y}{3}\\\dfrac{x}{4}=\dfrac{y}{2}+1\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}3x+3y=5x-5y\\x=2y+4\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}2x-8y=0\\x-2y=4\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x-4y=0\\x-2y=4\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}y=2\\x=8\end{matrix}\right.\)

a: \(\left\{{}\begin{matrix}x+4y=-11\\5x-4y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=-10\\x+4y=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\y=\dfrac{-11-x}{4}=\dfrac{-11+\dfrac{5}{3}}{4}=-\dfrac{7}{3}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}2x-y=7\\3x+5y=-22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-3y=21\\6x+15y=-66\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-18y=78\\2x-y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-13}{3}\\x=\dfrac{y+7}{2}=\dfrac{4}{3}\end{matrix}\right.\)

Bài 1: ĐKXĐ: $2\leq x\leq 4$
PT $\Leftrightarrow (\sqrt{x-2}+\sqrt{4-x})^2=2$

$\Leftrightarrow 2+2\sqrt{(x-2)(4-x)}=2$
$\Leftrightarrow (x-2)(4-x)=0$

$\Leftrightarrow x-2=0$ hoặc $4-x=0$

$\Leftrightarrow x=2$ hoặc $x=4$ (tm)

Bài 2:
PT $\Leftrightarrow 4x^3(x-1)-3x^2(x-1)+6x(x-1)-4(x-1)=0$

$\Leftrightarrow (x-1)(4x^3-3x^2+6x-4)=0$
$\Leftrightarrow x=1$ hoặc $4x^3-3x^2+6x-4=0$

Với $4x^3-3x^2+6x-4=0(*)$

Đặt $x=t+\frac{1}{4}$ thì pt $(*)$ trở thành:
$4t^3+\frac{21}{4}t-\frac{21}{8}=0$

Đặt $t=m-\frac{7}{16m}$ thì pt trở thành:

$4m^3-\frac{343}{1024m^3}-\frac{21}{8}=0$
$\Leftrightarrow 4096m^6-2688m^3-343=0$

Coi đây là pt bậc 2 ẩn $m^3$ và giải ta thu được \(m=\frac{\sqrt[3]{49}}{4}\) hoặc \(m=\frac{-\sqrt[3]{7}}{4}\)

Khi đó ta thu được \(x=\frac{1}{4}(1-\sqrt[3]{7}+\sqrt[3]{49})\)

Nãy mình tìm được một cách giải tương tự cho câu 2.

PT \(\Leftrightarrow\left(x-1\right)\left(4x^3-3x^2+6x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x^3-3x^2+6x-4=0\left(1\right)\end{matrix}\right.\)

Vậy pt có 1 nghiệm bằng 1.

\(\left(1\right)\Rightarrow8x^3-6x^2+12x-8=0\)

\(\Leftrightarrow7x^3+x^3-6x^2+12x-8=0\)

\(\Leftrightarrow\left(x-2\right)^3=-7x^3\)

\(\Leftrightarrow x-2=-\sqrt[3]{7}x\)

\(\Leftrightarrow x=\dfrac{2}{1+\sqrt[3]{7}}\)

Vậy pt có nghiệm \(S=\left\{1;\dfrac{2}{1+\sqrt[3]{7}}\right\}\)

Lưu ý: Nghiệm của người kia hoàn toàn tương đồng với nghiệm của mình (\(\dfrac{2}{1+\sqrt[3]{7}}=\dfrac{1}{4}\left(1-\sqrt[3]{7}+\sqrt[3]{49}\right)\))

Ta có hệ \(\hept{\begin{cases}\left(4x^2+1\right)x+\left(y-3\right)\sqrt{5-2y}=0\left(1\right)\\4x^2+y^2+2\sqrt{3-4x}=7\left(2\right)\end{cases}}\)

ĐK \(\hept{\begin{cases}y\ge\frac{5}{2}\\x\le\frac{3}{4}\end{cases}}\)

Đặt \(\hept{\begin{cases}2x=a\\\sqrt{5-2y}=b\ge0\end{cases}\Rightarrow\hept{\begin{cases}4x^2=a^2\\5-2y=b^2\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}4x^2=a^2\\y-3=\frac{5-b^2}{2}-3=\frac{-1-b^2}{2}\end{cases}}\)

Thế vào (1) ta có \(\left(a^2+1\right)\frac{a}{2}+\frac{-1-b^2}{2}b=0\)

\(\Leftrightarrow\frac{a^3+a}{2}+\frac{-b^3-b}{2}=0\Leftrightarrow a^3-b^3+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)

\(\Leftrightarrow a=b\)vì \(a^2+ab+b^2+1>0\forall a,b\)

\(\Rightarrow2x=\sqrt{5-2y}\Rightarrow4x^2=5-2y\Rightarrow y=\frac{5-4x^2}{2}\)

Thế y vào (2) ta có \(4x^2+\left(\frac{5-4x^2}{2}\right)^2+2.\sqrt{3-4x}=7\)

\(\Leftrightarrow16x^2+\left(5-4x^2\right)^2+8\sqrt{3-4x}=28\)\(\Leftrightarrow16x^2+25-40x^2+16x^4+8\sqrt{3-4x}-28=0\)

\(\Leftrightarrow16x^4-24x^2+8\sqrt{3-4x}-3=0\)

\(\Leftrightarrow\left(16x^4-1\right)-\left(24x^2-6\right)+\left(8\sqrt{3-4x}-8\right)=0\)

\(\Leftrightarrow\left(4x^2-1\right)\left(4x^2+1\right)-6\left(4x^2-1\right)+\left(8\sqrt{3-4x}-8\right)=0\)

\(\Leftrightarrow\left(4x^2-1\right)\left(4x^2+1\right)-6\left(4x^2-1\right)+8.\frac{2-4x}{\sqrt{3-4x}+1}=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)\left(4x^2+1\right)-6\left(2x+1\right)\left(2x-1\right)-8.2.\frac{2x-1}{\sqrt{3-4x}+1}=0\)

\(\Leftrightarrow\left(2x-1\right)\left[\left(2x+1\right)\left(4x^2+1\right)-6\left(2x+1\right)-\frac{16.1}{\sqrt{3-4x}+1}\right]=0\)

\(\Leftrightarrow\left(2x-1\right)\left[\left(2x+1\right)\left(4x^2-5\right)-\frac{16}{\sqrt{3-4x}+1}\right]=0\)

\(\Leftrightarrow2x-1=0\)

Vì với \(y=\frac{5-4x^2}{2}\ge\frac{5}{2}\Rightarrow4x^2-5< 0\Rightarrow\left(2x+1\right)\left(4x^2-5\right)-\frac{16}{\sqrt{3-4x}+1}< 0\)

\(\Leftrightarrow x=\frac{1}{2}\Rightarrow y=\frac{5-4\left(\frac{1}{2}\right)^2}{2}=2\)

Vậy hệ có nghiệm \(\left(x;y\right)=\left(\frac{1}{2};2\right)\)