X/4 = y/5;y/9 = z/5 và x - 2y + 3z = 27
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a, y \(\times\) \(\dfrac{4}{3}\) = \(\dfrac{16}{9}\)
y = \(\dfrac{16}{9}\) : \(\dfrac{4}{3}\)
y = \(\dfrac{4}{3}\)
b, ( y - \(\dfrac{1}{2}\)) + 0,5 = \(\dfrac{3}{4}\)
y - 0,5 + 0,5 = \(\dfrac{3}{4}\)
y = \(\dfrac{3}{4}\)
c, \(\dfrac{4}{5}-\dfrac{2}{5}y\) = 0,2
0,8 - 0,4y = 0,2
0,4y = 0,8 - 0,2
0,4y = 0,6
y = 1,5
d, (y + \(\dfrac{3}{4}\)) \(\times\) \(\dfrac{5}{7}\) = \(\dfrac{10}{9}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{10}{9}\) : \(\dfrac{5}{7}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{14}{9}\)
y = \(\dfrac{14}{9}\) - \(\dfrac{3}{4}\)
y = \(\dfrac{29}{36}\)
e, y : \(\dfrac{5}{4}\) = \(\dfrac{9}{5}\) + \(\dfrac{1}{2}\)
y : \(\dfrac{5}{4}\) = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{8}\)
f, y \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\) \(\times\) y = \(\dfrac{4}{5}\)
y \(\times\) ( \(\dfrac{1}{2}+\dfrac{3}{2}\)) = \(\dfrac{4}{5}\)
2y = \(\dfrac{4}{5}\)
y = \(\dfrac{2}{5}\)
a) \(\dfrac{10^{12}+5^{11}.2^9-5^{13}.2^8}{4.5^5.10^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^2.5^5.2^6.5^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^8.5^{11}}\)
\(=\dfrac{\left(2^8.5^{11}\right)\left(2^4.5+2-5^2\right)}{2^8.5^{11}}\)
\(=2^4.5+2-5^2\)
\(=57\)
b) \(\dfrac{\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x-y\right)^2\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x^2+y^2-2xy\right)\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y^2+x^2-2xy\right)}\)
\(=5\left(x-y\right)^2-3\left(x-y\right)+4\)
c) \(\dfrac{\left(x+y\right)^5-2\left(x+y\right)^4+3\left(x+y\right)^3}{-5\left(x+y\right)^3}\)
\(=\dfrac{\left(x+y\right)^3\left[5\left(x+y\right)^2-2\left(x+y\right)+3\right]}{-5\left(x+y\right)^3}\)
\(=\dfrac{5\left(x+y\right)^2-2\left(x+y\right)+3}{-5}\)
B=-4x^5y+x^4y^3-3x^2y^3z^2+4x^5y-2y^4-x^4y-x^4y+3y^4+4y^2x^2z^2-y^4+\(\frac{1}{2}\)
=(-4x^5y+4x^5y)+x^4y^3-3x^2y^3z^2+(2y^4+3y^4-y^4)+(-x^4y-x^4y)+4y^2x^2z^2+\(\frac{1}{2}\)
=x^4y^3-3y^3z^2-2x^4y+4y^2x^2z^2+\(\frac{1}{2}\)
vd câu 1:
ta có x-y=4 =>x=4+y
ta có pt:
4+y/y-2=3/2
=>8+2y=3y-6
=>-y=-14
=>y=14
=>x=4+y=4+14=18
các bài khác cũng tương tự thôi bạn
\(A=\dfrac{x^{\dfrac{5}{4}}y+xy^{\dfrac{5}{4}}}{\sqrt[4]{x}+\sqrt[4]{y}}\\ =\dfrac{xy\left(x^{\dfrac{1}{4}}+y^{\dfrac{1}{4}}\right)}{x^{\dfrac{1}{4}}+y^{\dfrac{1}{4}}}\\ =xy\)
\(B=\left(\sqrt[7]{\dfrac{x}{y}\sqrt[5]{\dfrac{y}{x}}}\right)^{\dfrac{35}{4}}\\= \left(\sqrt[7]{\dfrac{x}{y}\cdot\left(\dfrac{x}{y}\right)^{-\dfrac{1}{5}}}\right)^{\dfrac{35}{4}}\\ =\left(\sqrt[7]{\left(\dfrac{x}{y}\right)^{\dfrac{4}{5}}}\right)^{\dfrac{35}{4}}\\ =\left[\left(\dfrac{x}{y}\right)^{\dfrac{4}{35}}\right]^{\dfrac{35}{4}}\\ =\left(\dfrac{x}{y}\right)^{\dfrac{4}{35}\cdot\dfrac{35}{4}}\\ =\left(\dfrac{x}{y}\right)^1\\ =\dfrac{x}{y}\)
\(\dfrac{x}{4}=\dfrac{y}{5}\Rightarrow\dfrac{x}{4}.\dfrac{1}{9}=\dfrac{y}{5}.\dfrac{1}{9}\Rightarrow\dfrac{x}{36}=\dfrac{y}{45}\left(1\right)\)
\(\dfrac{y}{9}=\dfrac{z}{5}\Rightarrow\dfrac{y}{9}.\dfrac{1}{5}=\dfrac{z}{5}.\dfrac{1}{5}\Rightarrow\dfrac{y}{45}=\dfrac{z}{25}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)\(\Rightarrow\dfrac{x}{36}=\dfrac{y}{45}=\dfrac{z}{25}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{36}=\dfrac{y}{45}=\dfrac{z}{25}=\dfrac{2y}{90}=\dfrac{3z}{75}=\dfrac{x-2y+3z}{36-90+75}=\dfrac{27}{21}=\dfrac{9}{7}\)
thay vào tìm x , y ,z