Cho bài toán : Tìm y, biết 12 : (y x 3) = 4
Bạn A
12 : (y x 3) = 4
y x 3 = 12:4
yx3 = 3
y=1
Bạn B
12 : (y x 3) = 4
(12:y) : 3 = 4
12 : y = 12
y=1
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1)
xy + x - 4y = 12
x + y(x - 4) = 12
y(x - 4) = 12 - x
\(y=\dfrac{-x+12}{x-4}\)
Vì \(x,y\inℕ\) nên
\(\left(-x+12\right)⋮\left(x-4\right)\)
\(\left(-x+12\right)-\left(x-4\right)⋮\left(x-4\right)\)
\(16⋮\left(x-4\right)\)
\(\left(x-4\right)\inƯ\left(16\right)\)
\(\left(x-4\right)\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16\right\}\)
\(x\in\left\{5;3;6;2;8;0;12;-4;20;-12\right\}\)
\(y\in\left\{\dfrac{-5+12}{5-4};\dfrac{-3+12}{3-4};\dfrac{-6+12}{6-4};\dfrac{-2+12}{2-4};\dfrac{-8+12}{8-4};\dfrac{-0+12}{0-4};\dfrac{-12+12}{12-4};\dfrac{4+12}{-4-4};\dfrac{-20+12}{20-4};\dfrac{12+12}{-12-4}\right\}\)
\(y\in\left\{7;-9;3;-5;1;-3;0;-2;-\dfrac{1}{2};-\dfrac{7}{5}\right\}\)
\(\left(x;y\right)\in\left\{\left(5;7\right);\left(3;-9\right);\left(6;3\right);\left(2;-5\right);\left(8;1\right);\left(0;-3\right);\left(12;0\right);\left(-4;-2\right);\left(20;-\dfrac{1}{2}\right);\left(-12;-\dfrac{7}{5}\right)\right\}\)
Mà \(x,y\inℕ\) nên các giá trị cần tìm là \(\left(x;y\right)\in\left\{\left(5;7\right);\left(6;3\right);\left(8;1\right);\left(12;0\right)\right\}\)
2)
(2x + 3)(y - 2) = 15
\(\left(2x+3\right)\inƯ\left(15\right)\)
\(\left(2x+3\right)\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)
Ta lập bảng
2x + 3 | 1 | -1 | 3 | -3 | 5 | -5 | 15 | -15 |
y - 2 | 15 | -15 | 5 | -5 | 3 | -3 | 1 | -1 |
(x; y) | (-1; 17) | (-2; -13) | (0; 7) | (-3; -3) | (1; 5) | (-4; -1) | (6; 3) | (-9; 1) |
Mà \(x,y\inℕ\) nên các giá trị cần tìm là \(\left(x;y\right)\in\left\{\left(0;7\right);\left(1;5\right);\left(6;3\right)\right\}\)
\(4.\left(3x+y\right)^2+\left(x+y\right)^2\)
\(=3x^2+6xy+y^2+x^2-2xy+y^2\)
\(=9x^2+6xy+y^2+x^2-2xy+y^2\)
\(=10x^2-4xy+2y^2\)
\(7.\left(x-4\right)^2+\left(x+4y\right)\)
\(=x^2-8x+16+x+4y\)
\(=x^2-7x+16+4y\)
\(10.\left(2x+7\right)^2+\left(-2x-3\right)^2\)
\(=4x^2+28x+49+4x^2+12x+9\)
\(=8x^2+40x+58\)
\(12.-\left(x+1\right)^2-\left(x-1\right)^2\)
\(=-\left(x^2+2x+1\right)-\left(x^2-2x+1\right)\)
\(=-x^2-2x-1+x^2+2x-1\)
\(=4x\)
\(5.-\left(x+5\right)^2-\left(x-3\right)^2\)
\(=-\left(x^2+10x+25\right)-\left(x^2-6x+9\right)\)
\(=-x^2-10-25+x^2+6x-9\)
\(=-16x-16\)
\(8.-\left(-2x+3\right)^2-\left(5x-3\right)^2\)
\(=4x^2+12x+9-25x^2+30x-9\)
\(=-21x^2+42x\)
\(11.-\left(2x-y\right)^2-\left(x+3y\right)^2\)
\(=-4x^2+4xy-y^2-\left(x^2+6xy+9y^2\right)\)
\(=-4x^2+4xy-y^2-x^2-6xy-9y^2\)
\(=-5x^2-2xy-10y^2\)
4: =9x^2+6xy+y^2+x^2-2xy+y^2
=10x^2+4xy+2y^2
5: =-x^2-10x-25-x^2+6x-9
=-4x-34
7; \(=x^2-8xy+16y^2+x+4y\)
10: \(=4x^2+28x+49+4x^2+12x+9\)
=8x^2+40x+58
11: =-4x^2+4xy-y^2-x^2-6xy-9y^2
=-5x^2-2xy-10y^2
\(a,\dfrac{12}{5}=\dfrac{x}{1,5}\Rightarrow x=\dfrac{12\cdot1,5}{5}=3,6\\ b,\dfrac{x}{5}=\dfrac{3}{20}\Rightarrow x=\dfrac{5\cdot3}{20}=\dfrac{3}{4}\\ c,\dfrac{4}{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{4\cdot9}{10}=\dfrac{18}{5}\\ d,\Rightarrow\dfrac{x}{15}=\dfrac{60}{x}\Rightarrow x^2=60\cdot15=900\Rightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\\ 2,\)
a, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x+y-z}{3+5-6}=\dfrac{8}{2}=4\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=20\\z=24\end{matrix}\right.\)
b, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x-y+z}{3-5+6}=\dfrac{-4}{4}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-6\end{matrix}\right.\)
c, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{2y}{10}=\dfrac{3z}{18}=\dfrac{x-2y+3z}{3-10+18}=\dfrac{-33}{11}=-3\\ \Rightarrow\left\{{}\begin{matrix}x=-9\\y=-15\\z=-18\end{matrix}\right.\)
d, Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=k\Rightarrow x=3k;y=5k;z=6k\)
\(x^2-4y^2+2z^2=-475\\ \Rightarrow9k^2-100k^2+72z^2=-475\\ \Rightarrow-19k^2=-475\\ \Rightarrow k^2=25\Rightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=15;y=25;z=30\\x=-15;y=-25;z=-30\end{matrix}\right.\)
a) => y+42+2y= -12-14+2y
y+2y-2y = -12-14-42
y= -68
b) => 15+y-5-5y= -12-5y
y-5y+5y= -12-15+5
y = -22
c) => 2y+5-8y+21= -3-5y-2
2y-8y+5y= -3-2-5-21
-y= -31=>y=31
d)=> -13+3y+23= -120+y
3y-y= -120+13-23
2y= -130=>y= -65
e) => -21+32+5y= 16+4y
5y-4y= 16+21-32
y= 5
bài 1
a)y-(-42-2y) = (-12) - 14 +2y
y +42 + 2y = -12 -14 +2y
3y + 42 = -26 +2y
y = -68
b)15-(-y+5)-5y=-(12+5y+2)
15+y-5-5y=-12-5y-2
10-4y=-14-5y
-4y+5y=-14-10=-24
c)2y-(-5+8y-21)=-3-(5y+2)
2y+5-8y+21=-3y-5y-2
-6y+26=-8y-2
-6y+8y=-2-26
2y=-28
y=-28/2=-14
Bài 1:
a)\(\begin{cases}\left(x-3\right)^2+\left(y+2\right)^2=0\\\begin{cases}\left(x-3\right)^2\ge0\\\left(y+2\right)^2\ge0\end{cases}\end{cases}\)
\(\Rightarrow\begin{cases}\left(x-3\right)^2=0\\\left(y+2\right)^2=0\end{cases}\)\(\Rightarrow\begin{cases}x=3\\y=-2\end{cases}\)
b) tương tự
b) (x-12+y)200+(x-4-y)200= 0
\(\begin{cases}\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}=0\\\begin{cases}\left(x-12+y\right)^{200}\ge0\\\left(x-4-y\right)^{200}\ge0\end{cases}\end{cases}\)
\(\Rightarrow\begin{cases}\left(x-12+y\right)^{200}=0\\\left(x-4-y\right)^{200}=0\end{cases}\)\(\Rightarrow\begin{cases}x-12+y=0\\x-4-y=0\end{cases}\)\(\Rightarrow\begin{cases}x+y=12\left(1\right)\\x-y=4\left(2\right)\end{cases}\)
Trừ theo vế của (1) và (2) ta được:
\(2y=8\Rightarrow y=4\)\(\Rightarrow\begin{cases}x+4=12\\x-4=4\end{cases}\)\(\Rightarrow x=8\)
Vậy x=8; y=4
Hai bạn đều đúng
Bạn B phải bỏ dấu ngoặc
(12:y):3 =4
Hai bạn A và B đều đúng