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Bài 2:
\(2^{91}\) và \(5^{35}\)
Ta có:
\(2^{91}=\left(2^{13}\right)^7\) \(=8192^7\)
\(5^{35}=\left(5^5\right)^7\) =\(3125^7\)
Vì 8192\(^7\) >3125\(^7\) nên \(2^{91}>5^{35}\)
Bài 3:
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{ab}{cd}\)
VT=\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
\(=\dfrac{a^2-2ab+b^2}{c^2-2cd+d^2}\)
Mới biết làm đến đó thôi à!
2)
\(2^{91}=2^{13.7}=\left(2^{13}\right)^7=8192^7\)
\(5^{35}=5^{5.7}=\left(5^5\right)^7=3125^7\)
Vì \(8192>3125\)
Nên \(8192^7>3125^7\)
Vậy \(2^{91}>2^{35}\)
bz-cy/a = cx- az /b = ay-bx /c => bxz-cxy / ax = cxy-azy / b = azy-bxz/c = bxz-cxy + cxy-azy+azy-bxz / a+b+c = 0/ a+b+c = 0
Suy ra : bz -cy/a = 0 => bz-cy=0 => bz = cy => z/c = b/y
cx-az/b = 0 => cx-az=0 => cx=az => x/a = z/c
ay-bx/c = 0 => ay-bx = 0 => ay=bx=> y/b = x/a
Vậy x/a=y/b=c/z
1. Ta có:
a) \(\left(x-2y\right)\left(3xy-2y+3x\right)\)
\(=x\left(3xy-2y+3x\right)-2y\left(3xy-2y+3x\right)\)
\(=3x^2y-2xy+3x^2-6xy^2+4y^2-6xy\)
\(=3x^2y-6xy^2+3x^2-8xy+4y^2\)
b) \(\left(x-1\right)\left(x-2\right)\left(x-3\right)=\left(x-1\right)\left[\left(x-2\right)\left(x-3\right)\right]\)
\(=\left(x-1\right)\left[x\left(x-3\right)-2\left(x-3\right)\right]\)
\(=\left(x-1\right)\left(x^2-3x-2x+6\right)\)
\(=\left(x-1\right)\left(x^2-5x+6\right)\)
\(=x\left(x^2-5x+6\right)-1\left(x^2-5x+6\right)\)
\(=x^3-5x^2+6x-x^2+5x-6\)
\(=x^3-6x^2+11x-6\)
Đặt x/a=y/b=z/c=k
=>x=ak; y=bk; z=ck
\(\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{a^2k^2+b^2k^2+c^2k^2}{a^4k^2+b^4k^2+c^4k^2}=\dfrac{1}{a^2+b^2+c^2}\)
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 !!!
Phương Ann Nhã Doanh đề bài khó wá Mashiro Shiina Đinh Đức Hùng
Nguyễn Huy Tú Lightning Farron Akai Haruma