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Ta có: \(\frac{6x}{11}=\frac{9y}{2}=\frac{18z}{5}\Leftrightarrow\frac{-18x}{-33}=\frac{18y}{4}=\frac{18z}{5}\)
Áp dụng t/c của dãy tỉ số bằng nhau ta có:
\(\frac{-18x}{-33}=\frac{18y}{4}=\frac{18z}{5}=\frac{18\left(-x+y+z\right)}{-33+4+5}=\frac{18\cdot\left(-120\right)}{-24}=90\)
Do đó:
\(\frac{-18x}{-33}=90\Leftrightarrow x=165\)
\(\frac{18y}{4}=90\Leftrightarrow y=20\)
\(\frac{18z}{5}=90\Leftrightarrow z=25\)
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\(A=\left(\dfrac{-3}{7}.x^3.y^2\right).\left(\dfrac{-7}{9}.y.z^2\right).\left(6.x.y\right)\)
\(A=\left(\dfrac{-3}{7}x^3y^2\right).\left(\dfrac{-7}{9}yz^2\right).6xy\)
\(A=\left(\dfrac{-3}{7}.\dfrac{-7}{9}.6\right).\left(x^3.x\right)\left(y^2.y.y\right).z^2\)
\(A=2x^4y^4z^2\)
\(B=-4.x.y^3\left(-x^2.y\right)^3.\left(-2.x.y.z^3\right)^2\)
\(B=\left[\left(-4\right).\left(-2\right)\right].\left(x.x^6.x^2\right)\left(y^3.y^3.y^2\right)\left(z^6\right)\)
\(B=8x^7y^{y^8}z^6\)
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a, 3.x2.y + M - x.y=10x2y - 2xy
(3 x2y-xy) +M= 10x2y -2xy
M=10x2y-2xy+( 3x2y -xy)
M=(10x2y+3x2y)-(2xy+xy)
M=13 x2y-3xy
b,(6xy-5y2)-N=x2-2xy+4 y2
N= 6xy -5y2-( x2-2xy+4y2)
N= 6xy -5y2-x2 +2xy -4y2
N= (6xy +2xy)- (5y2+4y2)-x2
N= 8xy -9y2-x2
hok tốt
boy with luv
kt
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Câu 1:
c: 2x=3y
nên x/3=y/2
=>x/9=y/6
5y=3z
nên y/3=z/5
=>y/6=z/10
=>x/9=y/6=z/10
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{10}=\dfrac{3x+3y-7z}{3\cdot9+3\cdot6-7\cdot10}=\dfrac{35}{-25}=-\dfrac{7}{5}\)
Do đó: x=-63/5; y=-42/5; z=-14
Bài 2:
Gọi ba số lần lượt là a,b,c
Theo đề, ta có: 4/3a=b=3/4c
\(\Leftrightarrow\dfrac{a}{\dfrac{3}{4}}=\dfrac{b}{1}=\dfrac{c}{\dfrac{4}{3}}\)
\(\Leftrightarrow\dfrac{a}{9}=\dfrac{b}{12}=\dfrac{c}{16}\)
Đặt \(\dfrac{a}{9}=\dfrac{b}{12}=\dfrac{c}{16}=k\)
=>a=9k; b=12k; c=16k
Theo đề, ta có: \(a^2+b^2+c^2=481\)
\(\Leftrightarrow81k^2+144k^2+256k^2=481\)
=>k2=1
Trường hợp 1: k=1
=>a=9; b=12; c=16
Trường hợp 2: k=-1
=>a=-9; b=-12; c=-16
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Cho \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)
\(\frac{3x-y+5z}{x+y+3z}=\frac{3.2k-3k+5.5k}{2k+3k+3.5k}=\frac{6k-3k+25k}{2k+3k+15k}=\frac{28k}{21k}=\frac{4}{3}\)
Kb với minh nha!
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\(\frac{2}{3}x=\frac{3}{4}y=\frac{5}{6}z\)
=> \(\frac{2}{3}x.\frac{1}{30}=\frac{3}{4}y.\frac{1}{30}=\frac{5}{6}z.\frac{1}{30}\)
=> \(\frac{x}{45}=\frac{y}{40}=\frac{z}{36}\)
\(\Rightarrow\frac{x^2}{2025}=\frac{y^2}{1600}=\frac{z^2}{1296}\)
Đến đây bạn tự làm tiếp
\(\frac{2x}{3}=\frac{3y}{4}=\frac{5z}{6}< =>\frac{2x}{90}=\frac{3y}{120}=\frac{5z}{180}< =>\frac{x}{45}=\frac{y}{40}=\frac{z}{36}\)
\(< =>\frac{x^2}{2025}=\frac{y^2}{1600}=\frac{z^2}{1296}\)
Theo tính chất của dãy tỉ số bằng nhau thì
\(\frac{x^2}{2025}=\frac{y^2}{1600}=\frac{z^2}{1296}=\frac{x^2+y^2+z^2}{2025+1600+1296}=\frac{724}{4921}\)
\(< =>\hept{\begin{cases}4921x^2=724.2025=1466100\\4921y^2=724.1600=1158400\\4921z=724.1296=938304\end{cases}}\)
\(< =>\hept{\begin{cases}x\approx\pm17\\y\approx\pm15\\z\approx\pm14\end{cases}}\)
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\(A\le\left|A\right|=\dfrac{\left|xy+yz+xz\right|}{\left|xyz\right|}\)
Áp dụng: \(\left|a+b+c\right|\le\left|a\right|+\left|b\right|+\left|c\right|\)
\(\left|A\right|\le\dfrac{\left|xy\right|+\left|yz\right|+\left|xz\right|}{\left|xyz\right|}=\dfrac{1}{\left|x\right|}+\dfrac{1}{\left|y\right|}+\dfrac{1}{\left|z\right|}\)
\(\le\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}=1\)
Ta có đpcm. Dấu "=" khi \(x=y=z=3\)
Thêm 1 hướng suy nghĩ khác
Ta có: \(\left|x\right|\ge3;\left|y\right|\ge3;\left|z\right|\ge3\)
\(\Rightarrow0< \dfrac{1}{\left|x\right|}\le\dfrac{1}{3};0< \dfrac{1}{\left|y\right|}\le\dfrac{1}{3};0< \dfrac{1}{\left|z\right|}\le\dfrac{1}{3}\)
Ta có:
\(A=\dfrac{xy+yz+zx}{xyz}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\le\dfrac{1}{\left|x\right|}+\dfrac{1}{\left|y\right|}+\dfrac{1}{\left|z\right|}\le\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}=1\)