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Đặt \(a=\dfrac{1}{x};b=\dfrac{1}{y};c=\dfrac{1}{z}\Rightarrow xyz=1\) và \(x;y;z>0\)
Gọi biểu thức cần tìm GTNN là P, ta có:
\(P=\dfrac{1}{\dfrac{1}{x^3}\left(\dfrac{1}{y}+\dfrac{1}{z}\right)}+\dfrac{1}{\dfrac{1}{y^3}\left(\dfrac{1}{z}+\dfrac{1}{x}\right)}+\dfrac{1}{\dfrac{1}{z^3}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)}\)
\(=\dfrac{x^3yz}{y+z}+\dfrac{y^3zx}{z+x}+\dfrac{z^3xy}{x+y}=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)
\(P\ge\dfrac{\left(x+y+z\right)^2}{y+z+z+x+x+y}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}\)
\(P_{min}=\dfrac{3}{2}\) khi \(x=y=z=1\) hay \(a=b=c=1\)
Đặt \(a = \frac{1}{x} ; b = \frac{1}{y} ; c = \frac{1}{z} \Rightarrow x y z = 1\) và \(x ; y ; z > 0\)
Gọi biểu thức cần tìm GTNN là P, ta có:
\(P = \frac{1}{\frac{1}{x^{3}} \left(\right. \frac{1}{y} + \frac{1}{z} \left.\right)} + \frac{1}{\frac{1}{y^{3}} \left(\right. \frac{1}{z} + \frac{1}{x} \left.\right)} + \frac{1}{\frac{1}{z^{3}} \left(\right. \frac{1}{x} + \frac{1}{y} \left.\right)}\)
\(= \frac{x^{3} y z}{y + z} + \frac{y^{3} z x}{z + x} + \frac{z^{3} x y}{x + y} = \frac{x^{2}}{y + z} + \frac{y^{2}}{z + x} + \frac{z^{2}}{x + y}\)
\(P \geq \frac{\left(\left(\right. x + y + z \left.\right)\right)^{2}}{y + z + z + x + x + y} = \frac{x + y + z}{2} \geq \frac{3 \sqrt[3]{x y z}}{2} = \frac{3}{2}\)
\(P_{m i n} = \frac{3}{2}\) khi \(x = y = z = 1\) hay \(a = b = c = 1\)
a.
\(A=\left(\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x\left(x-1\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}+\dfrac{x-2}{x}\right):\dfrac{x+1}{x}\)
\(=\left(\dfrac{x^2+x+1}{x}+\dfrac{x+2}{x}+\dfrac{x-2}{x}\right):\dfrac{x+1}{x}\)
\(=\left(\dfrac{x^2+3x+1}{x}\right).\dfrac{x}{x+1}\)
\(=\dfrac{x^2+3x+1}{x+1}\)
2.
\(x^3-4x^3+3x=0\Leftrightarrow x\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=1\left(loại\right)\\x=3\end{matrix}\right.\)
Với \(x=3\Rightarrow A=\dfrac{3^2+3.3+1}{3+1}=\dfrac{19}{4}\)
Bài 4:
a. Vì $\triangle ABC\sim \triangle A'B'C'$ nên:
$\frac{AB}{A'B'}=\frac{BC}{B'C'}=\frac{AC}{A'C'}(1)$ và $\widehat{ABC}=\widehat{A'B'C'}$
$\frac{DB}{DC}=\frac{D'B'}{D'C}$
$\Rightarrow \frac{BD}{BC}=\frac{D'B'}{B'C'}$
$\Rightarrow \frac{BD}{B'D'}=\frac{BC}{B'C'}(2)$
Từ $(1); (2)\Rightarrow \frac{BD}{B'D'}=\frac{BC}{B'C'}=\frac{AB}{A'B'}$
Xét tam giác $ABD$ và $A'B'D'$ có:
$\widehat{ABD}=\widehat{ABC}=\widehat{A'B'C'}=\widehat{A'B'D'}$
$\frac{AB}{A'B'}=\frac{BD}{B'D'}$
$\Rightarrow \triangle ABD\sim \triangle A'B'D'$ (c.g.c)
b.
Từ tam giác đồng dạng phần a và (1) suy ra:
$\frac{AD}{A'D'}=\frac{AB}{A'B'}=\frac{BC}{B'C'}$
$\Rightarrow AD.B'C'=BC.A'D'$
14:
a: \(\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}=\frac{7x^2-22x+3}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{2x\left(5-3x\right)}{2x\left(x-3\right)\left(x+3\right)}=\frac{10x-6x^2}{2x\left(x-3\right)\left(x+3\right)}\)
b: \(\frac{x+1}{x-x^2}=\frac{-\left(x+1\right)}{x^2-x}=\frac{-\left(x+1\right)}{x\left(x-1\right)}=\frac{-\left(x+1\right)\cdot2\left(x-1\right)}{2x\left(x-1\right)^2}=\frac{-2x^2+2}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2x^2-4x+2}=\frac{x+2}{2\left(x^2-2x+1\right)}=\frac{x+2}{2\left(x-1\right)^2}=\frac{x\left(x+2\right)}{2x\left(x-1\right)^2}=\frac{x^2+2x}{2x\left(x-1\right)^2}\)
c: \(\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\cdot\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{2x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{6x^2+6x+6}{\left(x-1\right)\left(x_{}^2+x+1\right)}\)
d: \(\frac{7}{5x}=\frac{7\cdot2\cdot\left(x-2y\right)\left(x+2y\right)}{5x\cdot2\cdot\left(x-2y\right)\left(x+2y\right)}=\frac{14\left(x^2-4y^2\right)}{10x\left(x-2y\right)\left(x+2y\right)}=\frac{14x^2-56y^2}{10x\left(x-2y\right)\left(x+2y\right)}\)
\(\frac{4}{x-2y}=\frac{4\cdot5x\cdot2\cdot\left(x+2y\right)}{\left(x-2y\right)\cdot5x\cdot2\cdot\left(x+2y\right)}=\frac{40x\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}=\frac{40x^2+80xy}{10x\left(x-2y\right)\left(x+2y\right)}\)
\(\frac{y-x}{8y^2-2x^2}=\frac{x-y}{2x^2-8y^2}=\frac{x-y}{2\left(x-2y\right)\left(x+2y\right)}=\frac{5x\left(x-y\right)}{2\cdot5x\left(x-2y\right)\left(x+2y\right)}=\frac{5x^2-5xy}{10x\left(x-2y\right)\left(x+2y\right)}\)
e: \(\frac{5x^2}{x^3+6x^2+12x+8}=\frac{5x^2}{\left(x+2\right)^3}=\frac{5x^2\cdot2}{2\left(x+2\right)^3}=\frac{10x^2}{2\left(x+2\right)^3}\)
\(\frac{4x}{x^2+4x+4}=\frac{4x}{\left(x+2\right)^2}=\frac{4x\cdot2\cdot\left(x+2\right)}{2\left(x+2\right)^3}=\frac{8x^2+16x}{2\left(x+2\right)^3}\)
\(\frac{3}{2x+4}=\frac{3}{2\left(x+2\right)}=\frac{3\left(x+2\right)^2}{2\left(x+2\right)^3}=\frac{3\left(x^2+4x+4\right)}{2\left(x+2\right)^3}=\frac{3x^2+12x+12}{2\left(x+2\right)^3}\)
13:
a: \(\frac{25}{14x^2y}=\frac{25\cdot3\cdot y^4}{14x^2y\cdot3y^4}=\frac{75y^4}{45x^2y^5}\)
\(\frac{14}{21xy^5}=\frac{14\cdot2\cdot x}{2x\cdot21xy^5}=\frac{28x}{42x^2y^5}\)
b: \(\frac{11}{102x^4y}=\frac{11\cdot y^2}{102x^4y\cdot y^2}=\frac{11y^2}{102x^4y^3}\)
\(\frac{3}{34xy^3}=\frac{3\cdot x^3\cdot3}{34xy^3\cdot3x^3}=\frac{9x^3}{102x^4y^3}\)
c: \(\frac{3x+1}{12xy^4}=\frac{\left(3x+1\right)\cdot3\cdot x}{12xy^4\cdot3x}=\frac{9x^2+3x}{36x^2y^4}\)
\(\frac{y-2}{9x^2y^3}=\frac{\left(y-2\right)\cdot4\cdot y}{9x^2y^3\cdot4y}=\frac{4y^2-8y}{36x^2y^4}\)
d: \(\frac{1}{6x^3y^2}=\frac{1\cdot6\cdot xy^2}{6x^3y^2\cdot6xy^2}=\frac{6xy^2}{36x^4y^4}\)
\(\frac{x+1}{9x^2y^4}=\frac{\left(x+1\right)\cdot4\cdot x^2}{9x^2y^4\cdot4x^2}=\frac{4x^3+4x^2}{36x^4y^4}\)
\(\frac{x-1}{4xy^3}=\frac{\left(x-1\right)\cdot9\cdot x^3y}{4xy^3\cdot9x^3y}=\frac{9x^4y-9x^3y}{36x^4y^4}\)
e: \(\frac{3+2x}{10x^4y}=\frac{\left(2x+3\right)\cdot4y^4}{10x^4y\cdot4y^4}=\frac{8xy^4+12y^4}{40x^4y^5}=\frac{3\left(8xy^4+12y^4\right)}{3\cdot40x^4y^4}=\frac{24xy^4+36y^4}{120x^4y^4}\)
\(\frac{5}{8x^2y^2}=\frac{5\cdot5\cdot x^2y^3}{8x^2y^2\cdot5x^2y^3}=\frac{25x^2y^3}{40x^4y^5}=\frac{25x^2y^3\cdot3}{40x^4y^5\cdot3}=\frac{75x^2y^3}{120x^4y^5}\)
\(\frac{2}{3xy^5}=\frac{2\cdot40\cdot x^3}{3xy^5\cdot40x^3}=\frac{80x^3}{120x^4y^5}\)
f: \(\frac{4x-4}{2x\left(x+3\right)}=\frac{2\cdot\left(x-1\right)}{2x\cdot\left(x+3\right)}=\frac{x-1}{x\left(x+3\right)}=\frac{\left(x-1\right)\cdot3\left(x+1\right)}{3x\left(x+3\right)\left(x+1\right)}=\frac{3x^2-3}{3x\left(x+3\right)\left(x+1\right)}\)
\(\frac{x-3}{3x\left(x+1\right)}=\frac{\left(x-3\right)\left(x+3\right)}{3x\left(x+1\right)\left(x+3\right)}=\frac{x^2-9}{3x\left(x+1\right)\left(x+3\right)}\)
g: \(\frac{2x}{\left(x+2\right)^3}=\frac{2x\cdot2x}{2x\left(x+2\right)^3}=\frac{4x^2}{2x\left(x+2\right)^3}\)
\(\frac{x-2}{2x\left(x+2\right)^2}=\frac{\left(x-2\right)\left(x+2\right)}{2x\left(x+2\right)^2\cdot\left(x+2\right)}=\frac{x^2-4}{2x\left(x+2\right)^3}\)
h: \(\frac{5}{3x^3-12x}=\frac{5}{3x\left(x^2-4\right)}=\frac{5}{3x\left(x-2\right)\left(x+2\right)}=\frac{5\cdot2\left(x+3\right)}{3x\left(x-2\right)\left(x+2\right)\cdot2\left(x+3\right)}=\frac{10x+30}{6x\left(x-2\right)\left(x+2\right)\left(x+3\right)}\)
\(\frac{3}{\left(2x+4\right)\left(x+3\right)}=\frac{3}{2\left(x+2\right)\left(x+3\right)}=\frac{3\cdot3x\left(x-2\right)}{2\left(x+2\right)\left(x+3\right)\cdot3x\left(x-2\right)}=\frac{9x^2-18x}{6x\left(x-2\right)\left(x+2\right)\left(x+3\right)}\)
a:
b: TH1: \(\hat{BAD}>90^0;\hat{ABD}>90^0\)
Ta có: ABCD là hình thang
=>\(\hat{ABC}+\hat{BCD}=180^0\)
=>\(\hat{BCD}<180^0-90^0=90^0\)
=>\(\hat{BCD}<\hat{BAD}\)
TH2: \(\hat{ADC}>90^0;\hat{DCB}>90^0\)
Ta có: ABCD là hình thang
DC//AB
=>\(\hat{CDA}+\hat{DAB}=180^0\)
=>\(\hat{DAB}<180^0-90^0=90^0\)
=>\(\hat{DAB}<\hat{DCB}\)
c: Xét tứ giác ABCD có
AB//CD
AB=CD
Do đó: ABCD là hình bình hành
a) ĐKXĐ:
x³ - 1 khác 0
x khác 1
b) A = (5x² + 5x + 5)/(x³ - 1)
= 5(x² + x + 1)/[(x - 1)(x² + x + 1)]
= 5/(x - 1)
Thay x = 7 vào A, ta được:
A = 5/(7 - 1)
= 5/6