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\(y^2=xz;x^2=yt\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{y}{z};\dfrac{x}{y}=\dfrac{t}{x}\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{t}{x}\)
Đặt:
\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{t}{x}=k\) \(\Rightarrow\left\{{}\begin{matrix}x=yk\\y=zk\\t=xk\end{matrix}\right.\)
Thay vào tính
Theo đề bài đã cho, ta có:
\(y^2\)=xz => \(\dfrac{x}{y}\)=\(\dfrac{y}{z}\) (1)
\(z^2\)=yt => \(\dfrac{y}{z}\)=\(\dfrac{z}{t}\)(2)
Từ (1) và (2) suy ra:
\(\dfrac{x}{y}\)=\(\dfrac{y}{z}\)=\(\dfrac{z}{t}\)=\(\dfrac{x^3}{y^3}\)=\(\dfrac{y^3}{z^3}\)=\(\dfrac{z^3}{t^3}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{x^3}{y^3}\)=\(\dfrac{y^3}{z^3}\)=\(\dfrac{z^3}{t^3}\)=\(\dfrac{x^3+y^3+z^3}{y^3+z^3+t^3}\)
Mặt khác\(\dfrac{x^3}{y^3}\)=\(\dfrac{y^3}{z^3}\) =\(\dfrac{z^3}{t^3}\)=\(\dfrac{x^3y^3z^3}{y^3z^3t^3}\)=\(\dfrac{x^3}{t^3}\)
Từđó ta suy ra \(\dfrac{x^3+y^3+z^3}{y^3+z^3+t^3}\)= \(\dfrac{x^3}{t^3}\)
( bạn ghi sai đề nên mk đã sửa lại )
Lời giải:
\(y^2=xz\Rightarrow \frac{y}{z}=\frac{x}{y}\)
\(z^2=yt\Rightarrow \frac{z}{t}=\frac{y}{z}\)
Vậy \(\frac{x}{y}=\frac{y}{z}=\frac{z}{t}\)
Ta có:
\(\frac{x}{y}=\frac{y}{z}=\frac{z}{x}\Rightarrow \frac{x^3}{y^3}=\frac{y^3}{z^3}=\frac{z^3}{t^3}=\frac{x^3+y^3+z^3}{y^3+z^3+t^3}(1)\) (áp dụng tính chất dãy tỉ số bằng nhau)
\(\frac{x}{y}=\frac{y}{z}=\frac{z}{x}\Rightarrow \frac{x^3}{y^3}=\frac{x}{y}.\frac{y}{z}.\frac{z}{t}=\frac{x}{t}(2)\)
Từ \((1);(2)\Rightarrow \frac{x^3+y^3+z^3}{y^3+z^3+t^3}=\frac{x}{t}\) (đpcm)
5a.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{19.21}\\ =\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}.\dfrac{20}{21}=\dfrac{10}{21}\)
b.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\\ =\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}.1=\dfrac{1}{2}\)
bài 1:
|x| = \(\dfrac{1}{3}\) => x = \(\pm\)\(\dfrac{1}{3}\) |y| = 1 => y = \(\pm\)1
a
+) A = 2x\(^2\) - 3x + 5
= 2\(\left(\dfrac{1}{3}\right)^2\) - 3.\(\dfrac{1}{3}\) +5 = 2.\(\dfrac{1}{9}\) - 1 + 5
= \(\dfrac{2}{9}\) - 1 + 5 = \(\dfrac{2-9+45}{9}\) = \(\dfrac{38}{9}\)
+) A = 2x\(^2\) - 3x + 5
= 2\(\left(\dfrac{-1}{3}\right)^2\) - 3\(\left(\dfrac{-1}{3}\right)\) + 5
= 2.\(\dfrac{1}{9}\) - (-1) + 5 = \(\dfrac{2}{9}\) + 1 +5
= \(\dfrac{2+9+45}{9}\) = \(\dfrac{56}{9}\)
b) +) B = 2x\(^2\) - 3xy + y\(^2\)
= 2\(\left(\dfrac{1}{3}\right)^2\) - 3.\(\dfrac{1}{3}\).1 + 1\(^2\)
= 2.\(\dfrac{1}{9}\) - 1 + 1 = \(\dfrac{2}{9}\) - 1 + 1
= \(\dfrac{2-9+9}{9}\) = \(\dfrac{2}{9}\)
+) B = 2x\(^2\) - 3xy + y\(^2\)
= 2\(\left(\dfrac{-1}{3}\right)\)\(^2\) - 3\(\left(\dfrac{-1}{3}\right)\). 1 + 1\(^2\)
= 2.\(\dfrac{1}{9}\) - (-1) + 1 = \(\dfrac{2}{9}\) + 1 + 1
= \(\dfrac{2+9+9}{9}\) = \(\dfrac{20}{9}\)
bài 3
x.y.z = 2 và x + y + z = 0
A = ( x + y )( y +z )( z + x )
= x + y . y + z . z + x = ( x + y + z ) + ( x . y . z )
= 0 + 2 = 2
bài 4
a) | 2x - \(\dfrac{1}{3}\) | - \(\dfrac{1}{3}\) = 0 => | 2x - \(\dfrac{1}{3}\) | = \(\dfrac{1}{3}\)
=> 2x - \(\dfrac{1}{3}\) = \(\pm\) \(\dfrac{1}{3}\)
+) 2x - \(\dfrac{1}{3}\)= \(\dfrac{1}{3}\)
=> 2x = \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) = \(\dfrac{2}{3}\)
x = \(\dfrac{2}{3}\) : 2 = \(\dfrac{2}{3}\) . \(\dfrac{1}{2}\) = \(\dfrac{1}{3}\)
+) 2x - \(\dfrac{1}{3}\) = \(\dfrac{-1}{3}\)
2x = \(\dfrac{-1}{3}\) + \(\dfrac{1}{3}\) = 0
x = 0 : 2 = 2
b: \(3x^2y^3=\dfrac{1}{9}\)
\(\Leftrightarrow3x^2=\dfrac{1}{9}:\dfrac{1}{27}=3\)
=>x=1 hoặc x=-1
a: \(A=\dfrac{-2}{3}\cdot\left(-27\right)\cdot4\cdot\dfrac{1}{2}=18\cdot2=36\)
I . Trắc Nghiệm
1B . 2D . 3C . 5A
II . Tự luận
2,a,Ta có: A+(x\(^2\)y-2xy\(^2\)+5xy+1)=-2x\(^2\)y+xy\(^2\)-xy-1
\(\Leftrightarrow\) A=(-2x\(^2\)y+xy\(^2\)-xy-1) - (x\(^2\)y-2xy\(^2\)+5xy+1)
=-2x\(^2\)y+xy\(^2\)-xy-1 - x\(^2\)y+2xy\(^2\)-5xy-1
=(-2x\(^2\)y - x\(^2\)y) + (xy\(^2\)+ 2xy\(^2\)) + (-xy - 5xy ) + (-1 - 1)
= -3x\(^2\)y + 3xy\(^2\) - 6xy - 2
b, thay x=1,y=2 vào đa thức A
Ta có A= -3x\(^2\)y + 3xy\(^2\) - 6xy - 2
= -3 . 1\(^2\) . 2 + 3 .1 . 2\(^2\) - 6 . 1 . 2 -2
= -6 + 12 - 12 - 2
= -8
3,Sắp xếp
f(x) =9-x\(^5\)+4x-2x\(^3\)+x\(^2\)-7x\(^4\)
=9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x
g(x) = x\(^5\)-9+2x\(^2\)+7x\(^4\)+2x\(^3\)-3x
=-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x
b,f(x) + g(x)=(9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x) + (-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x)
=9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x
=(9-9)+(-x\(^5\)+x\(^5\))+(-7x\(^4\)+7x\(^4\))+(-2x\(^3\)+2x\(^3\))+(x\(^2\)+2x\(^2\))+(4x-3x)
= 3x\(^2\) + x
g(x)-f(x)=(-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x) - (9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x)
=-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x-9+x\(^5\)+7x\(^4\)+2x \(^3\)-x\(^2\)-4x
=(-9-9)+(x\(^5\)+x\(^5\))+(7x\(^4\)+7x\(^4\))+(2x\(^3\)+2x\(^3\))+(2x\(^2\)-x\(^2\))+(3x-4x)
= -18 + 2x\(^5\) + 14x\(^4\) + 4x\(^3\) + x\(^2\) - x
I . Trắc Nghiệm 1B . 2D . 3C . 5A II . Tự luận 2,a,Ta có: A+(x22y-2xy22+5xy+1)=-2x22y+xy22-xy-1 ⇔⇔ A=(-2x22y+xy22-xy-1) - (x22y-2xy22+5xy+1) =-2x22y+xy22-xy-1 - x22y+2xy22-5xy-1 =(-2x22y - x22y) + (xy22+ 2xy22) + (-xy - 5xy ) + (-1 - 1) = -3x22y + 3xy22 - 6xy - 2 b, thay x=1,y=2 vào đa thức A Ta có A= -3x22y + 3xy22 - 6xy - 2 = -3 . 122 . 2 + 3 .1 . 222 - 6 . 1 . 2 -2 = -6 + 12 - 12 - 2 = -8 3,Sắp xếp f(x) =9-x55+4x-2x33+x22-7x44 =9-x55-7x44-2x33+x22+4x g(x) = x55-9+2x22+7x44+2x33-3x =-9+x55+7x44+2x33+2x22-3x b,f(x) + g(x)=(9-x55-7x44-2x33+x22+4x) + (-9+x55+7x44+2x33+2x22-3x) =9-x55-7x44-2x33+x22+4x-9+x55+7x44+2x33+2x22-3x =(9-9)+(-x55+x55)+(-7x44+7x44)+(-2x33+2x33)+(x22+2x22)+(4x-3x) = 3x22 + x g(x)-f(x)=(-9+x55+7x44+2x33+2x22-3x) - (9-x55-7x44-2x33+x22+4x) =-9+x55+7x44+2x33+2x22-3x-9+x55+7x44+2x 33-x22-4x =(-9-9)+(x55+x55)+(7x44+7x44)+(2x33+2x33)+(2x22-x22)+(3x-4x) = -18 + 2x55 + 14x44 + 4x33 + x22 - x
1.
Đặt \(\dfrac{x}{5}=\dfrac{y}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\end{matrix}\right.\)
\(\Rightarrow x^2-y^2=\left(5k\right)^2-\left(4k\right)^2=25k^2-16k^2=9k^2=4\)
\(\Rightarrow k^2=\dfrac{4}{9}\Rightarrow k=\pm\dfrac{2}{3}\)
\(\circledast k=\dfrac{2}{3}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10}{3}\\y=\dfrac{8}{3}\end{matrix}\right.\)
\(\circledast k=-\dfrac{2}{3}\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{3}\\y=-\dfrac{8}{3}\end{matrix}\right.\)
2.
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+1+3y-2}{5+7}=\dfrac{2x+3y-1}{12}=\dfrac{2x+3y-1}{6x}\)
\(\Rightarrow6x=12\Rightarrow x=2\)
\(\Rightarrow y=\dfrac{\dfrac{2\cdot2+1}{5}\cdot7+2}{3}=3\)
3.
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Leftrightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-\left(z-3\right)}{4+9-4}=\dfrac{95-8+3}{9}=10\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10\cdot4+2}{2}=21\\y=\dfrac{10\cdot9+6}{3}=32\\z=10\cdot4+3=43\end{matrix}\right.\)
Bài2:
Vì x:y:z tỉ lệ với 4:5:6 =>\(\dfrac{x}{4}\)=\(\dfrac{y}{5}\)=\(\dfrac{z}{6}\) mà \(x^2\)-\(2y^2\)+\(z^2\)= 18
Ta có:
\(\dfrac{x}{4}\)=\(\dfrac{x^2}{16}\)
\(\dfrac{y}{5}\)=\(\dfrac{2y}{5}\)=\(\dfrac{2y^2}{10}\)
\(\dfrac{z}{6}\)=\(\dfrac{z^2}{36}\)
Áp dụng tính chất dãy tỉ số= nhau,ta có:
\(\dfrac{x^2}{16}\)=\(\dfrac{2y^2}{10}\)=\(\dfrac{z^2}{36}\)=\(\dfrac{x^2-2y^2+z^2}{16-10+36}\)=\(\dfrac{18}{42}\)=\(\dfrac{3}{7}\)
\(\dfrac{x^2}{16}\)=\(\dfrac{3}{7}\)
=> \(x^2\)=\(\dfrac{48}{7}\)
=> x=\(\sqrt{\dfrac{48}{7}}\)
\(\dfrac{2y^2}{10}\)=\(\dfrac{3}{7}\)
=> \(2y^2\)=\(\dfrac{30}{7}\)
2y=\(\sqrt{\dfrac{30}{7}}\)
y=\(\sqrt{\dfrac{30}{7}}\):2
y= 1,035098339.....
\(\dfrac{z^2}{36}\)=\(\dfrac{3}{7}\)
=> \(z^2\)=\(\dfrac{108}{7}\)
z= \(\sqrt{\dfrac{108}{7}}\)
Ta có :\(y^2=xz\Rightarrow\dfrac{x}{y}=\dfrac{y}{z}\)(1)
\(x^2=yt\Rightarrow\dfrac{x}{y}=\dfrac{t}{x}\) (2)
Từ (1) và (2) , ta suy ra :\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{t}{x}\)
Đặt \(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{t}{x}=k\)\(\)(3)
\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{t}{x}=k\Rightarrow k^3=\dfrac{x^3}{y^3}=\dfrac{y^3}{z^3}=\dfrac{t^3}{x^3}=\dfrac{x^3+y^3+t^3}{y^3+z^3+x^3}\)
\(\Rightarrow\dfrac{t^3}{x^3}=\dfrac{x^3+y^3+t^3}{y^3+z^3+x^3}\)
\(\Rightarrow\dfrac{x^3}{t^3}=\dfrac{x^3+y^3+z^3}{x^3+y^3+t^3}\)
\(\Rightarrow\dfrac{x^3+y^3+z^3}{x^3+y^3+t^3}=\left(\dfrac{x}{t}\right)^3\)
Đề có sai không vậy bạn
\(\left\{{}\begin{matrix}y^2=xz\\x^2=yt\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{y}{z}\\\dfrac{x}{y}=\dfrac{t}{x}\end{matrix}\right.\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{t}{x}\)
Đặt:
\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{t}{x}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=yk\\y=zk\\t=xk\end{matrix}\right.\)
Thay vào tính :v