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Bài 2:
a)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)
=> a = b = c
b)
\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{x}\)
=> x = y = z (theo a)
Thay x = y = z vào biểu thức, ta có:
\(M=\dfrac{x^{333}.x^{666}}{x^{999}}=1\)
c)
\(ac=b^2\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\)
\(ab=c^2\Rightarrow\dfrac{b}{c}=\dfrac{c}{a}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\Rightarrow a=b=c\)
Thay a = b = c vào biểu thức, ta có:
\(M=\dfrac{a^{333}}{a^{111}.a^{222}}=1\)
4.a
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Leftrightarrow\left(3x-y\right).4=3\left(x+y\right)\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow12x-3x=4y+3y\\ \Rightarrow9x=7y\\ \Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Từ \(a\left(y+z\right)=b\left(z+x\right)\), áp dụng t/c dãy tỉ số bằng nhau ta được
\(\dfrac{z+x}{a}=\dfrac{y+z}{b}=\dfrac{z+x-y-z}{a-b}=\dfrac{x-y}{a-b}\)
\(\Rightarrow\dfrac{z+x}{a}.\dfrac{1}{c}=\dfrac{y+z}{b}.\dfrac{1}{c}=\dfrac{x-y}{c\left(a-b\right)}\)(1)
Tương tự : từ \(b\left(z+x\right)=c\left(x+y\right)\)
\(\Rightarrow\dfrac{z+x}{c}=\dfrac{x+y}{b}=\dfrac{z+x-x-y}{c-b}=\dfrac{y-z}{c-b}\)\(\Rightarrow\dfrac{z+x}{c}.\dfrac{1}{a}=\dfrac{x+y}{b}.\dfrac{1}{a}=\dfrac{y-z}{c-b}.\dfrac{1}{a}\)
\(\Rightarrow\dfrac{z+x}{ac}=\dfrac{x+y}{ab}=\dfrac{y-z}{a\left(c-b\right)}\)(2)
từ \(a\left(y+z\right)=c\left(x+y\right)\)
\(\Rightarrow\dfrac{y+z}{c}=\dfrac{x+y}{a}=\dfrac{y+z-x-y}{c-a}=\dfrac{z-x}{c-a}\)\(\Rightarrow\dfrac{y+z}{c}.\dfrac{1}{b}=\dfrac{x+y}{a}.\dfrac{1}{b}=\dfrac{z-x}{c-a}.\dfrac{1}{b}\)
\(\Rightarrow\dfrac{y+z}{bc}=\dfrac{x+y}{ab}=\dfrac{z-x}{b\left(c-a\right)}\)(3)
Kết hợi (1);(2)(3) => ĐPCM
tik mik nha !!!
1)\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2017}{2018}\)
\(B=\dfrac{1}{2018}\)
2)a)\(x^2-2x-15=0\)
\(\Leftrightarrow x^2-2x+1-16=0\)
\(\Leftrightarrow\left(x-1\right)^2-16=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
3)\(\dfrac{a}{b}=\dfrac{d}{c}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a}{b}\cdot\dfrac{d}{c}=\dfrac{ad}{bc}\)
Lại có:\(\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a^2+d^2}{b^2+c^2}\)
\(\Rightarrow\dfrac{a^2+d^2}{b^2+c^2}=\dfrac{ad}{bc}\)
4)Ta có:\(g\left(x\right)=-x^{101}+x^{100}-x^{99}+...+x^2-x+1\)
\(g\left(x\right)=-x^{101}+\left(x^{100}-x^{99}+...+x^2-x+1\right)\)
\(g\left(x\right)=-x^{101}+f\left(x\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=f\left(x\right)+x^{101}-f\left(x\right)=x^{101}\)
Tại x=0 thì f(x)-g(x)=0
Tại x=1 thì f(x)-g(x)=1
Bài 1:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a, Ta có: \(\dfrac{a+c}{c}=\dfrac{bk+dk}{dk}=\dfrac{\left(b+d\right)k}{dk}=\dfrac{b+d}{d}\)
\(\Rightarrowđpcm\)
b, Ta có: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\) (1)
\(\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=\dfrac{k\left(b-d\right)}{b-d}=k\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
c, Ta có: \(\dfrac{a-c}{a}=\dfrac{bk-dk}{bk}=\dfrac{k\left(b-d\right)}{bk}=\dfrac{b-d}{b}\)
\(\Rightarrowđpcm\)
d, Ta có: \(\dfrac{3a+5b}{2a-7b}=\dfrac{3bk+5b}{2bk-7b}=\dfrac{b\left(3k+5\right)}{b\left(2k-7\right)}=\dfrac{3k+5}{2k-7}\)(1)
\(\dfrac{3c+5d}{2c-7d}=\dfrac{3dk+5d}{2dk-7d}=\dfrac{d\left(3k+5\right)}{d\left(2k-7\right)}=\dfrac{3k+5}{2k-7}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
e, Sai đề
f, \(\left(\dfrac{a-b}{c-d}\right)^{2012}=\left(\dfrac{bk-b}{dk-d}\right)^{2012}=\left[\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right]^{2012}=\dfrac{b^{2012}}{d^{2012}}\)(1)
\(\dfrac{a^{2012}+b^{2012}}{c^{2012}+d^{2012}}=\dfrac{b^{2012}k^{2012}+b^{2012}}{d^{2012}k^{2012}+d^{2012}}=\dfrac{b^{2012}\left(k^{2012}+1\right)}{d^{2012}\left(k^{2012}+1\right)}=\dfrac{b^{2012}}{d^{2012}}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
Sửa câu a:
(x - 2)2 - 36 = 0
(x - 2 - 6)(x - 2 + 6) = 0
(x - 8)(x + 4)= 0
\(\Leftrightarrow \begin{bmatrix} x - 8= 0 & & \\ x + 4 = 0 & & \end{bmatrix}\)
\(\Leftrightarrow \begin{bmatrix} x = 8 & & \\ x = - 4 & & \end{bmatrix}\)
pn bỏ dấu ngoặc bên phải nhé
Vậy x = 8; x = - 4
2:
\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)
\(\Rightarrow\dfrac{a+5}{b+6}=\dfrac{a-5}{b-6}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{a+5}{b+6}=\dfrac{a-5}{b-6}=\dfrac{a+5-a+5}{b+6-b+6}=\dfrac{10}{12}=\dfrac{5}{6}=\dfrac{a+5+a-5}{b+6+b-6}=\dfrac{2a}{2b}=\dfrac{a}{b}\)
Từ đó suy ra \(\dfrac{a}{b}=\dfrac{5}{6}\)
\(\RightarrowĐPCM\)
b/
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2b+c-a+2c-b+a+2a+b-c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
* \(\left\{{}\begin{matrix}2b+c-a=2a\\2c-b+a=2b\\2a+b-c=2c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2b+c=3a\\2c+a=3b\\2a+b=3c\end{matrix}\right.\)
+)\(\Rightarrow\left\{{}\begin{matrix}c=3a-2b\\a=3b-2c\\b=3c-2a\end{matrix}\right.\)
\(\Rightarrow\left(3a-2b\right)\left(3b-2c\right)\left(3c-2a\right)=abc\left(1\right)\)
+) \(\Rightarrow\left\{{}\begin{matrix}2b=3c-a\\2c=3b-a\\2a=3c-b\end{matrix}\right.\)
\(\Rightarrow\left(3a-c\right)\left(3b-a\right)\left(3c-b\right)=8abc\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{abc}{8abc}=\dfrac{1}{8}\)
\(\Rightarrow P=\dfrac{1}{8}\)
Theo tính chất của dãy tỉ số bằng nhau:
\(x=\dfrac{b+c}{a}=\dfrac{a+c}{b}=\dfrac{a+b}{c}=\dfrac{b+c+a+c+a+b}{a+b+c}=2\)
⇒ A = (x2 - x + 1)10 = (22 - 2 + 1)10 = 310