Vì (1/3-2x)^102 và (3y-x)104 lớn hơn hoặc bằng 0 với mọi x và y

=>(1/3-2x)^102 và (3y-x)^104=0

Ta có: (1/3-2x)^102=0

=>1/3-2x=0

=>2x=1/3

=>x=1/6

Ta có:(3y-x)^104=0

=>3y-1/6=0

=>3y=1/6

=>y=1/18

Vậy x=1/6 và y=1/18

Ta có:

\(\left(\frac{1}{3}-2x\right)^{102}+\left(3y-x\right)^{104}=0\left(1\right)\)

Nhận thấy:

\(\left(\frac{1}{3}-2x\right)^{102}\ge0;\left(3y-x\right)^{104}\ge0\forall x,y\)

Do đó (1) xảy ra khi và chỉ khi:

\(\hept{\begin{cases}\frac{1}{3}-2x=0\\3y-x=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{6}\\y=\frac{1}{18}\end{cases}}\)

Có sai không bạn

1) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y+z}{8-12+15}=\dfrac{10}{11}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{10}{11}\\\dfrac{y}{12}=\dfrac{10}{11}\\\dfrac{z}{15}=\dfrac{10}{11}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{80}{11}\\y=\dfrac{120}{11}\\z=\dfrac{150}{11}\end{matrix}\right.\)

2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{136}{62}=\dfrac{68}{31}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{68}{31}\\\dfrac{y}{20}=\dfrac{68}{31}\\\dfrac{z}{28}=\dfrac{68}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1020}{31}\\y=\dfrac{1360}{31}\\z=\dfrac{1904}{31}\end{matrix}\right.\)

3) \(\Rightarrow\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}\)

Áp dụng t/c dtsbn:

\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)

Gọi A là giao điểm \(k_3\) và \(k_4\Rightarrow\) tọa độ A là nghiệm:

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

\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\) \(\Rightarrow A\left(1;1\right)\)

3 đường thẳng đồng quy \(\Leftrightarrow d\) đi qua A

\(\Rightarrow\left(m^2-3m\right).1+2m-5=1\)

\(\Leftrightarrow m^2-m-6=0\Rightarrow\left[{}\begin{matrix}m=3\\m=-2\end{matrix}\right.\)

\(\frac{x}{2}=\frac{y}{3}\\ \Rightarrow\frac{x}{10}=\frac{y}{15}\)

\(\frac{y}{5}=\frac{z}{7}\\ \Rightarrow\frac{y}{10}=\frac{z}{14}\)

\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{14}=\frac{2x}{20}=\frac{3y}{45}=\frac{z}{14}=\frac{2x+3y+z}{20+45+14}=\frac{102}{79}\)

\(\Rightarrow x=\frac{1020}{79};y=\frac{1530}{79};z=\frac{1428}{79}\)

suy ra:  x/10 = y/15  ;   y/15 = z/21 và 2x +3y +z =102

suy ra: x/10 = y/15 = z/21 và  2x +3y +z =102

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

x/10=y/15=z/21 = 2x/20 = 3y/45 = z/21 = 2x+3y+z / 20 +45 +21 = 102/86 = 51/43 

SAI ĐỀ RÙI

\(a,\Leftrightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Leftrightarrow5\left(x+2\right)=0\Leftrightarrow x=-2\\ b,\Leftrightarrow2x\left(x-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ c,\Leftrightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Leftrightarrow3x\left(-x-2\right)=0\Leftrightarrow-3x\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

\(a,\left(2x-1\right)^2-\left(x-3\right)\left(x+3\right)-1969\\ =4x^2-4x+1-x^2+9-1969\\ =3x^2-4x-1959\)

\(b,\left(2x-3y\right)\left(2x+3y\right)-\left(2x-y\right)^2\\ =4x^2-9y^2-4x^2+4xy-y^2\\ =8y^2+4xy=4y\left(2y+x\right)\)

\(c,\left(x+3y\right)^2+\left(x+y\right)\left(x-y\right)+280\\ =x^2+6xy+9y^2+x^2-y^2+280\\ =2x^2+8y^2+6xy+280\)

a: \(\left(2x-1\right)^2-\left(x-3\right)\cdot\left(x+3\right)-1969\)

\(=4x^2-4x+1-x^2+9-1969\)

\(=3x^2-4x-1959\)

b: \(\left(2x-3y\right)\left(2x+3y\right)-\left(2x-y\right)^2\)

\(=4x^2-9y^2-4x^2+4xy-y^2\)

\(=-10y^2+4xy\)

a)\(\text{( 2 x − 1 )^2− ( x − 3 ) ( x + 3 ) − 1969}\)

\(\text{= 4x^2 − 4x + 1 − x^2 + 9 − 1969}\)

\(\text{=3x^2− 4 x − 1959}\)

b) \(\text{( 2 x − 3 y ) ( 2 x + 3 y ) − ( 2 x − y )^2}\)

=\(\text{= 4 x^2− 9 y^2− 4 x^2 + 4 x y − y^2}\)

\(\text{= -10 y^2+ 4 x y = -2 y ( 5 y -2 x )}\)

c)\(\text{( x + 3 y )^2 + ( x + y ) ( x − y ) + 280}\)

\(\text{= x^2 + 6 x y + 9 y^2 + x^2 − y^2 + 280}\)

\(\text{= 2 x^2 + 8 y^2 + 6 x y + 280}\)