Lời giải:
Áp dụng TCDTSBN:

$\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}$

$=\frac{4(3x-2y)}{16}=\frac{3(2z-4x)}{9}=\frac{2(4y-3z)}{4}$

$=\frac{4(3x-2y)+3(2z-4x)+2(4y-3z)}{16+9+4}=\frac{0}{29}=0$

$\Rightarrow 3x-2y=2z-4x=4y-3z=0$

$\Rightarrow 3x=2y; 2z=4x$

$\Rightarrow \frac{x}{2}=\frac{y}{3}; \frac{x}{2}=\frac{z}{4}$

$\Rightarrow \frac{x}{2}=\frac{y}{3}=\frac{z}{4}$

Đặt $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=t$

$\Rightarrow x=2t; y=3t; z=4t$. Khi đó:

$x+2y+3z=20$

$\Rightarrow 2t+2.3t+3.4t=20$

$\Rightarrow 2t+6t+12t=20$
$\Rightarrow 20t=20\Rightarrow t=1$

Do đó:

$x=2t=2; y=3t=3; z=4t=4$

@Nguyễn Chí Dũng, bạn muốn like hay là tick xanh?

cho mink huỷ câu trả lời ở dưới nhé!

\(\left\{{}\begin{matrix}mx-y=2m\\4x-my=m+6\end{matrix}\right.\)

Để hpt vô nghiệm thì: \(\dfrac{m}{4}=\dfrac{-1}{-m}\ne\dfrac{2m}{m+6}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{m}{4}=\dfrac{1}{m}\\\dfrac{1}{m}\ne\dfrac{2m}{m+6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}m^2=4\\2m^2-m-6\ne0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}m=\pm2\\m\ne2;m\ne-\dfrac{3}{2}\end{matrix}\right.\\ \Leftrightarrow m=-2\)

Để hpt có vô số nghiệm thì: 

\(\dfrac{m}{4}=\dfrac{-1}{-m}=\dfrac{2m}{m+6}\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{m}{4}=\dfrac{1}{m}\\\dfrac{1}{m}=\dfrac{2m}{m+6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}m^2=4\\2m^2-m-6=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}m=\pm2\\m=-\dfrac{3}{2};m=2\end{matrix}\right.\\ \Leftrightarrow m=2\) 

Để hpt có nghiệm thì \(\dfrac{m}{4}\ne\dfrac{1}{m}\Leftrightarrow m\ne\pm2\) 

\(\left\{{}\begin{matrix}mx-y=2m\\4x-my=m+6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}m^2x-my=2m^2\\4x-my=m+6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}mx-y=2m\\\left(m^2-4\right)x=2m^2-m-6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}m\cdot\dfrac{2m+3}{m+2}-y=2m\\x=\dfrac{2m^2-m-6}{m^2-4}=\dfrac{\left(2m+3\right)\left(m-2\right)}{\left(m+2\right)\left(m-2\right)}=\dfrac{2m+3}{m+2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2m^2+3m}{m+2}-2m\\x=\dfrac{2m+3}{m+2}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2m^2+3m-2m\left(m+2\right)}{m+2}\\x=\dfrac{2m+3}{m+2}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2m+3m-2m^2-4m}{m+2}=\dfrac{-m}{m+2}\\x=\dfrac{2m+3}{m+2}\end{matrix}\right.\)

1.

$-4x^2-4x+18=0$

$\Leftrightarrow 4x^2+4x-18=0$

$\Leftrightarrow (2x+1)^2-19=0$

$\Leftrightarrow (2x+1)^2=19$

$\Rightarow 2x+1=\pm \sqrt{19}$

$\Rightarrow x=\frac{\pm \sqrt{19}-1}{2}$

2.

$18x^2+114x+330=0$

$\Leftrightarrow 9x^2+57x+165=0$

$\Leftrightarrow (3x+9,5)^2+74,75=0$

$\Leftrightarrow (3x+9,5)^2=-74,75<0$ (vô lý) 

Do đó PT vô nghiệm.

9

thank

5)

a) \(3x+8y=26\)

 \(\Leftrightarrow y=\dfrac{26-3x}{8}\)

 Vì \(y\inℤ\) nên \(\dfrac{26-3x}{8}\inℤ\)

 \(\Rightarrow26-3x⋮8\)

 \(\Leftrightarrow3x\equiv2\left(mod8\right)\)

 Vì \(ƯCLN\left(3,8\right)=1\) nên đặt \(x=8q+r\left(0\le r< 8\right)\) thì:

 \(3\left(8q+r\right)\equiv2\left(mod8\right)\)

 \(\Leftrightarrow24q+3r\equiv2\left(mod8\right)\)

 \(\Leftrightarrow3r\equiv2\left(mod8\right)\)

 Thử từng trường hợp, ta thấy ngay \(r=6\).

 Vậy \(x=8q+6\) 

\(\Rightarrow y=\dfrac{26-3x}{8}=\dfrac{26-3\left(8q+6\right)}{8}=\dfrac{8-24q}{8}=1-3q\)

 Vậy phương trình đã cho có nghiệm nguyên là \(\left(8q+6,1-3q\right)\) với \(q\inℤ\) bất kì.

 b) Cho \(1-3q>0\Leftrightarrow q< \dfrac{1}{3}\) 

 Cho \(8q+6>0\Leftrightarrow q>-\dfrac{3}{4}\)

 Do đó \(-\dfrac{3}{4}< q< \dfrac{1}{3}\). Mà \(q\inℤ\Rightarrow q=0\)

 Thế vào \(x,y\), pt sẽ có nghiệm nguyên dương là \(\left(6;1\right)\)

 Câu 6 làm tương tự nhé bạn.

(30 + 30) -( 60 + 60)

= 60 - 120 

= 60 + (-120)

= - 60

-60

\(100+12-30-1-2-3\)

\(=100+\left(12-2\right)-30-1-3\)

\(=100+10-30-1-3\)

\(=110-30-1-3\)

\(=80-1-3\)

\(=79-3\)

\(=76\)

76

\(60^3=60.60.60=3600.60=216000\)

216000

81

\(9^2=9.9=81\)