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NV
Tìm x,y,z biết 6x 4z 5 2y 5x 6 5z 6y 4và 3x 2y 5z 96 tìm x,y,z biết 6x 4z 5 2y 5x 6 5z 6y 4 và 3x 2y
0
27 tháng 6 2023
A=5x^2+6x^2+3y+7y=11x^2+10y
B=7x^3+6x^3+6y+5y+36=13x^3+11y+36
C=-8x^5-x^5+3y^4-10y^4=-9x^5-7y^4
C=x^2-5x^2+y^2-6y^2=-4x^2-5y^2
NH
2
11 tháng 10 2023
\(2x^2+6x-4\left(x+3\right)\)
\(=\left(2x^2+6x\right)-4\left(x+3\right)\)
\(=2x\left(x+3\right)-4\left(x+3\right)\)
\(=\left(x+3\right)\left(2x+4\right)\)
\(=2\left(x+3\right)\left(x+2\right)\)
______
\(xy\left(x-y\right)-5x+5y\)
\(=xy\left(x-y\right)-\left(5x-5y\right)\)
\(=xy\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-5\right)\)
______
\(2x^2+3x-4xy-6y\)
\(=\left(2x^2+3x\right)-\left(4xy+6y\right)\)
\(=x\left(2x+3\right)-2y\left(2x+3\right)\)
\(=\left(x-2y\right)\left(2x+3\right)\)
KV
11 tháng 10 2023
2x² + 6x - 4(x + 3)
= (2x² + 6x) - 4(x + 3)
= 2x(x + 3) - 4(x + 3)
= (x + 3)(2x - 4)
= 2(x + 3)(x - 2)
------------
xy(x - y) - 5x + 5y
= xy(x - y) - (5x - 5y)
= xy(x - y) - 5(x - y)
= (x - y)(xy - 5)
------------
2x² + 3x - 4xy - 6y
= (2x² - 4xy) + (3x - 6y)
= 2x(x - 2y) + 3(x - 2y)
= (x - 2y)(2x + 3)
6 tháng 1 2021
a) Ta có: \(Q=-x^2-y^2+4x-4y+2=-\left(x^2+y^2-4x+4y-2\right)\)
\(=-\left(x^2-4x+4+y^2+4y+4\right)+10\)
\(=-\left[\left(x-2\right)^2+\left(y+2\right)^2\right]+10\le10\forall x,y\)
Vậy MaxQ=10 khi x=2, y=-2
b) +Ta có: \(A=-x^2-6x+5=-\left(x^2+6x-5\right)=-\left(x^2+6x+9-14\right)\)
\(=-\left(x^2+6x+9\right)+14=-\left(x+3\right)^2+14\le14\forall x\)
Vậy MaxA=14 khi x=-3
+Ta có: \(B=-4x^2-9y^2-4x+6y+3=-\left(4x^2+9y^2+4x-6y-3\right)\)
\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)
\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2\right]+5\le5\forall x,y\)
Vậy MaxB=5 khi x=-1/2, y=1/3
c) Ta có: \(P=x^2+y^2-2x+6y+12=x^2-2x+1+y^2+6y+9+2\)
\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\forall x,y\)
Vậy MinP=2 khi x=1, y=-3
4 tháng 1 2022
ĐKXĐ: \(x,y\ne0\)
Hệ pt \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6x+6y}{xy}=5\\\dfrac{4}{x}+\dfrac{3}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{x}+\dfrac{6}{y}=5\\\dfrac{4}{x}+\dfrac{3}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{x}+\dfrac{6}{y}=5\\\dfrac{3}{y}=1-\dfrac{4}{x}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{x}+2.\left(1-\dfrac{4}{x}\right)=5\\\dfrac{3}{y}=1-\dfrac{4}{x}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{x}+2-\dfrac{8}{x}=5\\\dfrac{3}{y}=1-\dfrac{4}{x}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-2}{x}=3\\\dfrac{3}{y}=1-\dfrac{4}{x}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-2}{3}\\y=\dfrac{3}{7}\end{matrix}\right.\)(tm)
Vậy...
24 tháng 6 2021
\(\dfrac{5z-6y}{4}=\dfrac{6x-4z}{5}=\dfrac{4y-5x}{6}\)
\(\Leftrightarrow\dfrac{4\left(5z-6y\right)}{16}=\dfrac{5\left(6x-4z\right)}{25}=\dfrac{6\left(4y-5x\right)}{36}\)
\(\Leftrightarrow\dfrac{20z-24y}{16}=\dfrac{30x-20z}{25}=\dfrac{24y-30x}{36}\)
ADTCDTSBN có:
\(\dfrac{20z-24y}{16}=\dfrac{30x-20z}{25}=\dfrac{24y-30x}{36}=\dfrac{20z-24y+30x-20z+24y-30x}{16+25+36}=0\)
Do đó \(20z-24y=0;30x-20z=0\)
\(\Leftrightarrow5z=6y;6x=4z\)
\(\Rightarrow y=\dfrac{5z}{6};x=\dfrac{4z}{6}\)
Có \(3x-3y+5z=96\Rightarrow3.\dfrac{4z}{6}-3.\dfrac{5z}{6}+5z=96\)
\(\Rightarrow z=\dfrac{64}{3}\) \(\Rightarrow y=\dfrac{160}{9}\)và \(x=\dfrac{128}{9}\)
Vậy...
25 tháng 6 2021
cho x/3 = y/4 và y/5 = z/6. tìm M = 2x + 3y+ 4z / 3x + 4y + 5z
22 tháng 5 2021
\(\left\{{}\begin{matrix}6x+6y=5xy(1)\\\dfrac{4}{x}-\dfrac{3}{y}=1\end{matrix}\right.\)
Chia 2 vế cho xy thì (1)(vì `x,y ne 0`)
`<=>` $\begin{cases}\dfrac6x+\dfrac6y=5\\\dfrac{4}{x}-\dfrac{3}{y}=1\\\end{cases}$
`<=>` $\begin{cases}\dfrac6x+\dfrac6y=5\\\dfrac{8}{x}-\dfrac{6}{y}=2\\\end{cases}$
`<=>` $\begin{cases}\dfrac{14}{x}=7\\\dfrac6x+\dfrac6y=5\\\end{cases}$
`<=>` $\begin{cases}\dfrac{14}{x}=7\\\dfrac6x+\dfrac6y=5\\\end{cases}$
`<=>` $\begin{cases}x=2\\y=3\\\end{cases}$
Vậy HPT có nghiệm (x,y)=(2,3)
22 tháng 5 2021
\(\left\{{}\begin{matrix}3x-3y=5\\5x+2y=23\end{matrix}\right.< =>\left\{{}\begin{matrix}6x-6y=10\\15x+6y=69\end{matrix}\right.< =>\left\{{}\begin{matrix}21x=79\\3x-3y=5\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x=\dfrac{79}{21}\\y=\dfrac{44}{21}\end{matrix}\right.\)
vậy hệ pt có nghiệm (x,y)=(\(\dfrac{79}{21};\dfrac{44}{21}\))
6^x:6^y=216*6^4
6^(x-y)=6^3*6^4
6^(x-y)=6^(3+4)
6^(x-y)=6^7
x-y=7
vậy x-y=7
\(6^x\div6^y=216.6^4\)
\(\Leftrightarrow6^{x-y}=6^3.6^4\)
\(\Leftrightarrow x-y=3+4=7\)