Câu hỏi của Trịnh Minh Châu - Toán lớp 8 - Học toán với OnlineMath

Tại sao lại có a?

tại đề bài gốc có a. ai biết được

a) \(\frac{8xy}{3x-1}:\frac{12xy^3}{5-15x}\)

\(=\frac{8xy}{3x-1}.\frac{5-15x}{12xy^3}\)

\(=\frac{2}{3x-1}.\frac{5\left(1-3x\right)}{3y^2}\)

\(=\frac{10}{3y^2}\)

b) \(\frac{2x+1}{x-2}:\left(-\frac{2x-1}{x-2}\right)\)

\(=\frac{2x+1}{x-2}.\frac{x-2}{1-2x}=\frac{2x+1}{1-2x}\)

=.=, làm nhanh lẫn

a) \(=\frac{8xy}{3x-1}.\frac{5\left(3x-1\right)}{-12xy^3}\)

\(=\frac{-10}{3y^2}\)

Cô si hết lên!

\(a+1\ge2\sqrt{a}\)

\(b+1\ge2\sqrt{b}\)

\(c+1\ge2\sqrt{c}\)

\(\Rightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge8\sqrt{abc}=8\)

Dấu "=" khi a = b = c = 1

Áp dụng bdt AM-GM cho 2 số dương a và 1 ta được:

\(a+1\ge2\sqrt{a}\)

tương tự ta có: \(b+1\ge2\sqrt{b}\);\(c+1\ge2\sqrt{c}\)

Suy ra \(\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge8\sqrt{abc}=8\)(do \(abc=1\Rightarrow\sqrt{abc}=1\))

Dấu = xảy ra \(\Leftrightarrow a=b=c=1\)(đpcm)

Ta có: \(x^2y-xy^2+y^2z-yz^2+xz^2-x^2z=xy\left(x-y\right)-z\left(x^2-y^2\right)+z^2\left(x-y\right)\)

\(=xy\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+z^2\left(x-y\right)=\left(x-y\right)\left(xy-zx-zy+z^2\right)\)

\(=\left(x-y\right)\left(x\left(y-z\right)-z\left(y-z\right)\right)=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)

\(\frac{x^2+6x+9}{\left(x-1\right)^2}.\frac{2x^2-4x-2}{4x^2+24x+36}\)

\(=\frac{x^2+6x+9}{\left(x-1\right)^2}.\frac{2x^2-4x-2}{4\left(x^2+6x+9\right)}\)

\(=\frac{1}{\left(x-1\right)^2}.\frac{2x^2-4x-2}{4}\)

\(=\frac{2x^2-4x-2}{4x^2-8x+4}\)

\(\frac{x^2+6x+9}{\left(x-1\right)^2}.\frac{2x^2-4x-2}{4x^2+24x+36}\)

\(=\frac{x^2+2\left(x\right)\left(3\right)+3^2}{\left(x-1\right)^2}.\frac{2x^2-4x-2}{4x^2+24x+36}\)

\(=\frac{\left(x+3\right)^2}{\left(x-1\right)^2}.\frac{2x^2+4x-2}{4x^2+24x+36}\)

\(=\frac{\left(x+3\right)^2}{\left(x-1\right)^2}.\frac{2\left(x^2-2x-1\right)}{4x^2+24x+36}\)

\(=\frac{\left(x+3\right)^2}{\left(x-1\right)^2}.\frac{2\left(x^2-2x-1\right)}{4\left(x^2+2\left(x\right)\left(3\right)+3^2\right)}\)

\(=\frac{1}{\left(x-1\right)^2}.\frac{2\left(x^2-2x-1\right)}{4}\)

\(=\frac{1.2\left(x^2-2x-1\right)}{\left(x-1\right)^2.4}\)

\(=\frac{2\left(x^2-2x-1\right)}{4\left(x-1\right)^2}\)

\(=\frac{x^2-2x-1}{2\left(x-1\right)^2}\)

\(\left(ac+bd\right)^2=a^2c^2+2abcd+b^2d^2\)

\(\left(ad-bc\right)^2=a^2d^2-2abcd+b^2c^2\)

\(\Rightarrow\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)

Mà \(\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)

Nên \(\left(ac+bd\right)^2+\left(ad-bc\right)^2=\left(a^2+b^2\right)\left(c^2+d^2\right)\)(đpcm)

Ta có \(VP=\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)(1)

\(VT=\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)

\(=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)(2)

từ (1),(2)\(\Rightarrowđpcm\)

Mình nghĩ đề như này

\(\frac{7x+2}{5xy^3}.\frac{x^2y^3}{21x+6}\)

\(=\frac{7x+2}{5}.\frac{x}{3\left(7x+2\right)}\)

\(=\frac{x}{15}\)

Hì...sorry tui bấm thiếu.

Cảm ơn bạn

Trong 3 ngày liên tiếp tui sẽ k cho bạn

\(2=x+y\ge2\sqrt{xy}\)(cô - si)

\(\Rightarrow\sqrt{xy}\le1\Rightarrow xy\le1\)

Ta có \(S=x^2+y^2=\left(x+y\right)^2-2xy\)

\(=4-2xy\ge4-2=2\)

Dấu "=" khi x = y = 1

Ta có: \(\left(x-y\right)^2\ge0\)\(\Leftrightarrow x^2-2xy+y^2\ge0\)

\(\Leftrightarrow x^2+y^2\ge2xy\)\(\Leftrightarrow2\left(x^2+y^2\right)\ge x^2+y^2+2xy\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)\(\Leftrightarrow x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)

Thay \(x+y=2\)vào bất phương trình ta được:\(x^2+y^2\ge\frac{2^2}{2}=\frac{4}{2}=2\)

Dấu " = " xảy ra \(\Leftrightarrow x-y=0\)\(\Leftrightarrow x=y\)

mà \(x+y=2\)\(\Rightarrow x=y=1\)

Vậy \(minS=2\)\(\Leftrightarrow x=y=1\)