a) \(2^x+124=5^7\)

\(2^x=78125-124\)

\(2^x=78001\)

b) \(10^x+528=7^2\)

\(10^x=49-528\)

\(10^x=-479\)

Trả lời:

Áp dụng Bất Đẳng Thức Cauchy-Schwarz ta có:

(3+1)(3x2+y2)≥(3x+y)2

⇒4(3x2+y2)≥(3x+y)2⇒4(3x2+y2)≥(3x+y)2

⇒4(3x2+y2)≥(3x+y)2=12=1⇒4(3x2+y2)≥(3x+y)2=12=1

⇒M=3x2+y2≥14⇒M=3x2+y2≥14

Đẳng thức xảy ra khi x=y=14

Ta có:  x + y = 1 => y = 1 - x

Khi đó: P = \(x^3+y^3+2x^2y^2=\left(x+y\right)^3-3xy\left(x+y\right)+2\left(xy\right)^2\)

\(=2\left(xy\right)^2-3xy+1=2\left[\left(xy\right)^2-2.xy.\frac{3}{4}+\frac{9}{16}\right]-\frac{1}{8}\)

\(=2\left(xy-\frac{3}{4}\right)^2-\frac{1}{8}\)

\(=2\left[x\left(1-x\right)-\frac{3}{4}\right]^2-\frac{1}{8}\)

\(=2\left[-x^2+x-\frac{3}{4}\right]^2-\frac{1}{8}\)

\(=2\left[\left(x-\frac{1}{2}\right)^2+\frac{1}{2}\right]^2-\frac{1}{8}\ge\frac{3}{8}\)

Dấu "=" xảy ra <=> x = y =1/2

Thiếu: 

Kết luận: Vậy GTNN của P = 3/8 đạt tại x = y = 1/2.

\(\frac{3x-2}{8}=\frac{5y+6}{3}=\frac{3x-5y-8}{8-3}=\frac{3x-5y-8}{5}\)

\(+,3x=5y+8\Rightarrow\frac{5y+6}{8}=\frac{5y+6}{3}\Rightarrow y=-\frac{6}{5}\Rightarrow x=\frac{2}{3}\)

\(+,3x\ne5y+8\Rightarrow5=10x\Leftrightarrow x=\frac{1}{2}\Rightarrow\frac{-1}{16}=\frac{5y+6}{3}\Rightarrow....\)

\(xy+x+y+1=5\Leftrightarrow x\left(y+1\right)+\left(y+1\right)=5\Leftrightarrow\left(x+1\right)\left(y+1\right)=5\)

x;y nguyên nên đến đây dễ rồi

\(2.3^{x-2}+4.3^{x-2}=162\Rightarrow6.3^{x-2}=162\Leftrightarrow3^{x-2}=\frac{162}{6}=27\)

\(\Rightarrow3^{x-2}=27=3^3\)

\(\Rightarrow x-2=3\Leftrightarrow x=5\)

Đáp án A

a) \(x-2xy+x=0\Leftrightarrow2x-2xy=0\)

\(\Leftrightarrow2x\left(1-y\right)=0\Leftrightarrow\hept{\begin{cases}2x=0\\1-y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=1\end{cases}}\)

b) \(2xy+x-2y=4\)

\(\Leftrightarrow x\left(2y+1\right)-2y-1=3\)

\(\Leftrightarrow x\left(2y+1\right)-\left(2y+1\right)=3\)

\(\Leftrightarrow\left(2y+1\right)\left(x-1\right)=3\)

Đến đây bí =) Alibaba!

\(a)x-2xy+x=0\)

\(\Leftrightarrow2x-2xy=0\)

\(\Leftrightarrow2x\left(1-y\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\1-y=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\y=1\end{cases}}}\)

Vậy \(x=0;y=1\)

a, 3x ( y+1) + y + 1 = 7

(y+1)(3x +1) =7

th1 : \(\left\{{}\begin{matrix}y+1=1\\3x+1=7\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=0\\x=2\end{matrix}\right.\)

th2: \(\left\{{}\begin{matrix}y+1=-1\\3x+1=-7\end{matrix}\right.\)=> x = -8/3 (loại)

th3: \(\left\{{}\begin{matrix}y+1=7\\3x+1=1\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}y=6\\x=0\end{matrix}\right.\)

th 4 : \(\left\{{}\begin{matrix}y+1=-7\\3x+1=-1\end{matrix}\right.\)=> x=-2/3 (loại)

Vậy (x,y)= (2 ;0);  (0; 6)

b, xy - x + 3y - 3 = 5

   (x( y-1) + 3( y-1) = 5

          (y-1)(x+3) = 5

 th1: \(\left\{{}\begin{matrix}y-1=1\\x+3=5\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=2\\x=8\end{matrix}\right.\)

th2: \(\left\{{}\begin{matrix}y-1=-1\\x+3=-5\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=0\\x=-8\end{matrix}\right.\)

th3: \(\left\{{}\begin{matrix}y-1=5\\x+3=1\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=6\\x=-2\end{matrix}\right.\)

th4: \(\left\{{}\begin{matrix}y-1=-5\\x+3=-1\end{matrix}\right.\) =>  \(\left\{{}\begin{matrix}y=-4\\x=-4\end{matrix}\right.\)

vậy (x, y) = ( 8; 2); ( -8; 0);  (-2; 6); (-4; -4)

c, 2xy + x + y = 7 => y = \(\dfrac{7-x}{2x+1}\) ; y ϵ Z ⇔ 7-x ⋮ 2x+1

⇔ 14 - 2x ⋮ 2x + 1 ⇔ 15 - 2x - 1  ⋮ 2x + 1

th1 : 2x + 1 = -1=> x = -1; y = \(\dfrac{7-(-1)}{-1.2+1}\) = -8

th2: 2x+ 1 = 1=> x =0; y = 7

th3: 2x+1 = -3 => x =  x=-2  => y = \(\dfrac{7-(-2)}{-2.2+1}\) = -3 

th4: 2x+ 1 = 3 => x = 1 => y = \(\dfrac{7+1}{2.1+1}\) = 2

th5: 2x + 1 = -5 => x = -3=> y = \(\dfrac{7-(-3)}{-3.2+1}\) = -2

th6: 2x + 1 = 5 => x = 2; ; y = \(\dfrac{7-2}{2.2+1}\) =1

th7 : 2x + 1 = -15 => x = -8; y = \(\dfrac{7-(-8)}{-8.2+1}\) = -1

th8 : 2x+1 = 15 => x = 7; y = \(\dfrac{7-7}{2.7+1}\) = 0

kết luận

(x,y) = (-1; -8); (0 ;7); ( -2; -3) ; ( 1; 2); ( -3; -2); (2;1); (-8;-1);(7;0)

3xy−2x+5y=293xy−2x+5y=29

9xy−6x+15y=879xy−6x+15y=87

(9xy−6x)+(15y−10)=77(9xy−6x)+(15y−10)=77

3x(3y−2)+5(3y−2)=773x(3y−2)+5(3y−2)=77

(3y−2)(3x+5)=77(3y−2)(3x+5)=77

⇒(3y−2)⇒(3y−2) và (3x+5)(3x+5) là Ư(77)=±1,±7,±11,±77Ư(77)=±1,±7,±11,±77

Ta có bảng giá trị sau:

Do x,y∈Zx,y∈Z nên (x,y)∈{(−4;−3),(−2;−25),(2;3),(24;1)}