tìm x
a. \(\sqrt{x^6+1}\)= x^ 2
b. (2x+1)^5= (2x-1)^2011
c. 12-|x-3|=5x+8
d. 2. (1/4x-3) - 5/6. (6x-3/5) = -2
tim x y z
\(\dfrac{7x-3y+12}{2y}\) = \(\dfrac{y+2z}{z-y+2}\)= \(\dfrac{x}{-y}\)
help me soon^-^
tìm x;y;z
a) \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\)
b) \(\dfrac{1+3y}{12}=\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}\)
c) \(\dfrac{3x}{8}=\dfrac{3y}{64}=\dfrac{3z}{216}\) và \(2x^2+2y^2-z^2\)
d) \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\) và \(x^2+y^2+z^2=14\)
Tìm x,y,z trong dãy tỉ số bằng nhau
1)\(\dfrac{3x}{8}=\dfrac{3y}{64}=\dfrac{3z}{216}\)và \(2x^2+2y^2.z^2=1\)
2) \(\dfrac{2x+1}{5}=\dfrac{4y-5}{9}=\dfrac{2x+4y-4}{7x}\)
3) \(\dfrac{x^3+y^3}{6}=\dfrac{x^3-2y^3}{4}\)và x6 . y6 =14
4) \(\dfrac{x+4}{6}=\dfrac{3y-1}{8}=\dfrac{3y-x-5}{x}\)
5) \(\dfrac{3}{x-1}=\dfrac{4}{y-2}=\dfrac{5}{z-3}\)và x.y.z=192
6)\(\dfrac{x-y}{3}=\dfrac{x+y}{13}=\dfrac{x.y}{200}\)
7)\(\dfrac{x+1}{2}=\dfrac{y-1}{3}=\dfrac{z+2}{4}=\dfrac{x+y+z+2}{2x+5}\)
8) \(\dfrac{15}{x-9}=\dfrac{20}{y-12}=\dfrac{40}{z-24}\)và x.y = 1200
9)\(\dfrac{40}{x-30}=\dfrac{20}{y-15}=\dfrac{28}{z-21}\) và x.y.z = 22400
10)15x = -10y =6z và x.y.z = -30000
11) Cho\(\dfrac{x+1}{3}=\dfrac{y-2}{5}=\dfrac{2z+14}{9}\)và x+z=y
12) Cho \(\dfrac{x}{3}=\dfrac{y}{4}\)và \(\dfrac{y}{5}=\dfrac{z}{6}\).Tính M=\(\dfrac{2x+3y+4z}{3x+4y+5z}\)
A = \(\dfrac{5xy^2-3z}{3xy}+\dfrac{4x^2y+3z}{3xy}\)
B = \(\dfrac{3y+5}{y-1}+\dfrac{-y^2-4y}{1-y}+\dfrac{y^2+y+7}{y-1}\)
C = \(\dfrac{6x}{x^2-9}+\dfrac{5x}{x-3}+\dfrac{x}{x+3}\)
D = \(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x-4x^2}\)
E = \(\dfrac{x^3+2x}{x^3+1}+\dfrac{2x}{x^2-x+1}+\dfrac{1}{x+1}\)
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
Tìm x và y biết:
a) \(\dfrac{2x}{3y}=\dfrac{-1}{3}\)và -2x+3y=7
b)\(\dfrac{x}{3}=\dfrac{y}{4}\)và 2x+5y=10
c) 7x=3y và x-y=16
d)\(\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{z}{-2}\)và x+y-2z=160
e)\(\dfrac{x}{10}=\dfrac{y}{5};\dfrac{y}{2}=\dfrac{z}{3}\)và 2x-3y+4z=330
f) \(\dfrac{x}{-2}=\dfrac{-y}{4}=\dfrac{z}{5}\)và x-2y+3z=1200
g) \(\dfrac{x}{3}=\dfrac{y}{8}=\dfrac{z}{5}\)và 2x+3y-z=50
a,
\(\dfrac{2x}{3y}=\dfrac{-1}{3}\\ \Rightarrow\dfrac{2x}{-1}=\dfrac{3y}{3}\\ \Leftrightarrow\dfrac{-2x}{1}=\dfrac{3y}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{-2x}{1}=\dfrac{3y}{3}=\dfrac{-2x+3y}{1+3}=\dfrac{7}{4}\)
\(\dfrac{-2x}{1}=\dfrac{7}{4}\Rightarrow-2x=\dfrac{7}{4}\Rightarrow x=\dfrac{7}{4}:\left(-2\right)=\dfrac{-7}{8}\\ \dfrac{3y}{3}=\dfrac{7}{4}\Rightarrow y=\dfrac{7}{4}\)
Vậy \(x=\dfrac{-7}{8};y=\dfrac{7}{4}\)
b,
\(\dfrac{x}{3}=\dfrac{y}{4}\\ \Leftrightarrow\dfrac{2x}{6}=\dfrac{5y}{20}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{6}=\dfrac{5y}{20}=\dfrac{2x+5y}{6+20}=\dfrac{10}{26}=\dfrac{5}{13}\\ \dfrac{x}{3}=\dfrac{2x}{6}=\dfrac{5}{13}\Rightarrow x=\dfrac{5}{13}\cdot3=\dfrac{15}{13}\\ \dfrac{y}{4}=\dfrac{5y}{20}=\dfrac{5}{13}\Rightarrow y=\dfrac{5}{13}\cdot4=\dfrac{20}{13}\)
Vậy \(x=\dfrac{15}{13};y=\dfrac{20}{13}\)
c,
\(7x=3y\\ \Rightarrow\dfrac{x}{3}=\dfrac{y}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{16}{-4}=-4\\ \dfrac{x}{3}=-4\Rightarrow x=\left(-4\right)\cdot3=-12\\ \dfrac{y}{7}=-4\Rightarrow y=\left(-4\right)\cdot7=-28\)
Vậy \(x=-12;y=-28\)
d,
\(\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{z}{-2}\\ \Leftrightarrow\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{-2z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{-2z}{4}=\dfrac{x+y+\left(-2z\right)}{5+1+4}=\dfrac{x+y-2z}{10}=\dfrac{160}{10}=16\\ \dfrac{x}{5}=16\Rightarrow x=16\cdot5=80\\ \dfrac{y}{1}=16\Rightarrow y=16\\ \dfrac{z}{-2}=\dfrac{-2z}{4}=16\Rightarrow z=16\cdot\left(-2\right)=-32\)
Vậy \(x=80;y=16;z=-32\)
e,
\(\dfrac{x}{10}=\dfrac{y}{5}\Rightarrow\dfrac{x}{20}=\dfrac{y}{10};\dfrac{y}{2}=\dfrac{z}{3}\Rightarrow\dfrac{y}{10}=\dfrac{z}{15}\\ \Rightarrow\dfrac{x}{20}=\dfrac{y}{10}=\dfrac{z}{15}\\ \Leftrightarrow\dfrac{2x}{40}=\dfrac{3y}{30}=\dfrac{4z}{60}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{40}=\dfrac{3y}{30}=\dfrac{4z}{60}=\dfrac{2x-3y+4z}{40-30+60}=\dfrac{330}{70}=\dfrac{33}{7}\)
\(\dfrac{x}{20}=\dfrac{2x}{40}=\dfrac{33}{7}\Rightarrow x=\dfrac{33}{7}\cdot20=\dfrac{660}{7}\\ \dfrac{y}{10}=\dfrac{3y}{30}=\dfrac{33}{7}\Rightarrow y=\dfrac{33}{7}\cdot10=\dfrac{330}{7}\\ \dfrac{z}{15}=\dfrac{4z}{60}=\dfrac{33}{7}\Rightarrow z=\dfrac{33}{7}\cdot15=\dfrac{495}{7}\)
Vậy \(x=\dfrac{660}{7};y=\dfrac{330}{7};z=\dfrac{495}{7}\)
f,
\(\dfrac{x}{-2}=\dfrac{-y}{4}=\dfrac{z}{5}\\ \Leftrightarrow\dfrac{x}{-2}=\dfrac{-2y}{8}=\dfrac{3z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{-2}=\dfrac{-2y}{8}=\dfrac{3z}{15}=\dfrac{x+\left(-2y\right)+3z}{\left(-2\right)+8+15}=\dfrac{x-2y+3z}{21}=\dfrac{1200}{21}=\dfrac{400}{7}\)
\(\dfrac{x}{-2}=\dfrac{400}{7}\Rightarrow x=\dfrac{400}{7}\cdot\left(-2\right)=\dfrac{-800}{7}\\ \dfrac{-y}{4}=\dfrac{-2y}{8}=\dfrac{400}{7}\Rightarrow-y=\dfrac{400}{7}\cdot4=\dfrac{1600}{7}\Rightarrow y=\dfrac{-1600}{7}\\ \dfrac{z}{5}=\dfrac{3z}{15}=\dfrac{400}{7}\Rightarrow z=\dfrac{400}{7}\cdot5=\dfrac{2000}{7}\)
Vậy \(x=\dfrac{-800}{7};y=\dfrac{-1600}{7};z=\dfrac{2000}{7}\)
g,
\(\dfrac{x}{3}=\dfrac{y}{8}=\dfrac{z}{5}\\ \Leftrightarrow\dfrac{2x}{6}=\dfrac{3y}{24}=\dfrac{z}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{6}=\dfrac{3y}{24}=\dfrac{z}{5}=\dfrac{2x+3y-z}{6+24-5}=\dfrac{50}{25}=2\)
\(\dfrac{x}{3}=\dfrac{2x}{6}=2\Rightarrow x=2\cdot3=6\\ \dfrac{y}{8}=\dfrac{3y}{24}=2\Rightarrow y=2\cdot8=16\\ \dfrac{z}{5}=2\Rightarrow z=2\cdot5=10\)
Vậy \(x=6;y=16;z=10\)
Làm gấp nên k có kiểm tra, bn bấm máy tính dò lại nhé
Bài 1 : Tìm x,y,z biết :
a) 2x = 3y ; 5y = 7z và 3x - 7y + 5z = -30
b) 3x =5y ; 7y = 2z và x + y + z = 74
c) x : z = \(\dfrac{2}{3}\) : \(\dfrac{1}{2}\) ; z : y = 1 : \(\dfrac{4}{7}\) và y + z = 66
d) x : y : z = 3 : 4 : 5 và \(2x^2\) + \(2y^2\) - \(3z^2\) = -100
e) \(x:y:z\) = 2 : 5 : 6 và \(2x^2\) + \(4y^2\) - \(4z^2\) = -324
f) \(\dfrac{x-1}{2}\) = \(\dfrac{y-2}{3}\) = \(\dfrac{z-3}{4}\) và \(x-2y+3z=14\)
g)\(\dfrac{x-1}{2}\) = \(\dfrac{y+3}{4}\) =\(\dfrac{z-5}{6}\) và \(5z-3x-4y=50\)
h) \(\dfrac{x}{2}=\dfrac{y}{7}\) và \(xy=56\)
i)\(\dfrac{x-y}{3}=\dfrac{x+y}{13}=\dfrac{xy}{200}\)
k) \(\dfrac{x-5}{6}=\dfrac{x+5}{18}\)
l) \(\dfrac{2x-11}{12}=\dfrac{x+5}{20}\)
5, Tìm x, y ϵ Z, sao cho:
a) y = \(\dfrac{6x-4}{2x+3}\) b) \(\dfrac{1}{x}-\dfrac{y}{2}=\dfrac{1}{4}\)
c) xy-3x+2y=5 d) (3x-5)(2x+1)=12
a) Để y nguyên thì \(6x-4⋮2x+3\)
\(\Leftrightarrow-13⋮2x+3\)
\(\Leftrightarrow2x+3\in\left\{1;-1;13;-13\right\}\)
\(\Leftrightarrow2x\in\left\{-2;-4;10;-16\right\}\)
hay \(x\in\left\{-1;-2;5;-8\right\}\)
BT10: Thực hiện phép tính
\(a,\dfrac{4}{5}y^2x^5-x^3.x^2y^2\)
\(b,-xy^3-\dfrac{2}{7}y^2.xy\)
\(c,\dfrac{5}{6}xy^2z-\dfrac{1}{4}xyz.y\)
\(d,15x^4+7x^4-20x^2.x^2\)
\(e,\dfrac{1}{2}x^5y-\dfrac{3}{4}x^5y+xy.x^4\)
\(f,13x^2y^5-2x^2y^5+x^6\)
a: =-1/5x^5y^2
b: =-9/7xy^3
c: =7/12xy^2z
d: =2x^4
e: =3/4x^5y
f: =11x^2y^5+x^6
a,\(\dfrac{x+1}{x-3}+\dfrac{-2x^2+2x}{x^2-9}+\dfrac{x-1}{x+3}\)
b,\(\dfrac{1-2x}{6x^3y}+\dfrac{3+2y}{6x^3y}+\dfrac{2x-4}{6x^3y}\)
c,\(\dfrac{5}{2x^2y}+\dfrac{3}{5xy^2}+\dfrac{x}{3y^3}\)
d,\(\dfrac{5}{4\left(x+2\right)}+\dfrac{8-x}{4x^2+8x}\)
c,\(\dfrac{x^2+2}{x^3+1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
Tính giá trị biểu thức sau:
a) A= 2x + 2y + 3xy(x+y) + 5(x3y2 + x2 y3) + 4 tại x+y = 0
b) B = \(\dfrac{3}{8}x^3y^2-\dfrac{1}{4}x^2y+xy^3\) tại x = -0,5 và y = 2
c) C = 4xyz( x-y-z) tại x = -1 , \(\left|y\right|\) = 2 , z = \(\dfrac{1}{3}\)
d) D = \(\dfrac{5x-3y}{2x-y}\) tại \(\dfrac{x}{y}=\dfrac{6}{5}\)
\(A=2x+2y+3xy\left(x+y\right)+5\left(x^3y^2+x^2y^3\right)\)
\(\Rightarrow A=2\left(x+y\right)+3xy\left(x+y\right)+5x^2y^2\left(x+y\right)\)
\(\Rightarrow A=0\) ( do x+y = 0 )