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NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 1 2024
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 \(x=3\Rightarrow A=\dfrac{3^2+3.3+1}{3+1}=\dfrac{19}{4}\)
AH
Akai Haruma
Giáo viên
3 tháng 2 2024
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:
$\frac{AD}{A'D'}=\frac{AB}{A'B'}=\frac{BC}{B'C'}$
$\Rightarrow AD.B'C'=BC.A'D'$
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 3 2023
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 \(a=b=c=1\)
Bài 1:
a) $x^2+6x+9$
$=x^2+2.x.3+3^2$
$=(x+3)^2$
b) $9x^2-6x+1$
$=(3x)^2-2.3x.1+1^2$
$=(3x-1)^2$
c) $x^2y^2+xy+\frac14$
$=(xy)^2+2.xy.\frac12+\left(\frac12\right)^2$
$=\left(xy+\frac12\right)^2$
d) $(x-y)^2+6(x-y)+9$
$=(x-y)^2+2.(x-y).3+3^2$
$=(x-y+3)^2$
Bài 2:
a) $-x^3+3x^2-3x+1$
$=1^3-3.1^2.x+3.1.x^2-x^3$
$=(1-x)^3$
b) $x^3+x^2+\frac13 x+\frac{1}{27}$
$=x^3+3.x^2.\frac13+3.x.\left(\frac13\right)^2+\left(\frac13\right)^3$
$=\left(x+\frac13\right)^3$
c) $x^6-3x^4y+3x^2y^2-y^3$
$=(x^2)^3-3.(x^2)^2.y+3.x^2.y^2-y^3$
$=(x^2-y)^3$
d) $(x-y)^3+(x-y)^2+\frac13 (x-y)+\frac{1}{27}$
$=(x-y)^3+3.(x-y)^2.\frac13+3.(x-y).\left(\frac13\right)^2+\left(\frac13\right)^3$
$=\left(x-y+\frac13\right)^3$
Bài 3:
a) $x^3+27$
$=x^3+3^3$
$=(x+3)(x^2-x.3+3^2)$
$=(x+3)(x^2-3x+9)$
b) $x^3-\frac18$
$=x^3-\left(\frac12\right)^3$
$=\left(x-\frac12\right)\left[x^2-x.\frac12+\left(\frac12\right)^2\right]$
$=\left(x-\frac12\right)\left(x^2-\frac12 x+\frac14\right)$
c) $8x^3+y^3$
$=(2x)^3+y^3$
$=(2x+y)[(2x)^2-2x.y+y^2]$
$=(2x+y)(4x^2-2xy+y^2)$
d) $8x^3-27y^3$
$=(2x)^3-(3y)^3$
$=(2x-3y)[(2x)^2+2x.3y+(3y)^2]$
$=(2x-3y)(4x^2+6xy+9y^2)$
Bài 4:
a) \(101^2=\left(100+1\right)^2\)
\(=100^2+2.100.1+1^2\)
\(=10000+200+1=10201\)
b) \(75^2-50.75+25^2\)
\(=75^2-2.75.25+25^2\)
\(=\left(75-25\right)^2\)
\(=50^2=2500\)
c) \(103.97\)
\(=\left(100+3\right).\left(100-3\right)\)
\(=100^2-3^2\\ =10000-9=9991\)
Bài 5:
a) \(\left(x+3y\right)^2-\left(x-3y\right)^2\)
\(=\left(x+3y-x+3y\right)\left(x+3y+x-3y\right)\\ =6y.2x=12xy\)
b) \(Q=\left(x-y\right)^2-4\left(x-y\right)\left(x+2y\right)+4\left(x+2y\right)^2\)
\(=\left(x-y\right)^2-2.\left(x-y\right).2\left(x+2y\right)+\left[2\left(x+2y\right)\right]^2\\ =\left[\left(x-y\right)-2\left(x+2y\right)\right]^2\\ =\left(x-y-2x-4y\right)^2\\ =\left(-x-5y\right)^2\)
c) \(A=\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+12\right)\)
\(=\left(x+2+x-2\right)\left[\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)+\left(x-2\right)^2\right]-2x\left(x^2+12\right)\\ =2x\left(x^2+4x+4-x^2+4+x^2-4x+4\right)-2x\left(x^2+12\right)\\ =2x\left(x^2+12\right)-2x\left(x^2+12\right)=0\)
d) \(B=\left(xy+2\right)^3-6\left(xy+2\right)^2+12\left(xy+2\right)-8\)
\(=\left(xy+2\right)^3-3.\left(xy+2\right)^2.2+3.\left(xy+2\right).2^2-2^3\\ =\left(xy+2-2\right)^3\\ =\left(xy\right)^3=x^3y^3\)
e) \(A=\left(x-3\right)\left(x^2+3x+9\right)-\left(x^3+3\right)\)
\(=\left(x-3\right)\left(x^2+x.3+3^2\right)-x^3-3\\ =x^3-3^3-x^3-3\\ =-27-3=-30\)
Bài 6:
\(a,VT=\left(a-b\right)^2=a^2-2ab+b^2\\ =\left(a^2+2ab+b^2\right)-4ab\\ =\left(a+b\right)^2-4ab=VP\\ b,VT=\left(x+y\right)^2+\left(x-y\right)^2\\ =x^2+2xy+y^2+x^2-2xy+y^2\\ =2x^2+2y^2\\ =2\left(x^2+y^2\right)=VP\\ c,VT=\left(x+y\right)^2-\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]\\ =\left(x+y-x+y\right)\left(x+y+x-y\right)\\ =2y.2x=4xy=VP\\ d,VT=\left(x-y\right)^2+\left(x+y\right)^2+2\left(x^2-y^2\right)\\ =\left(x-y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\\ =\left[\left(x-y\right)+\left(x+y\right)\right]^2\\ =\left(x-y+x+y\right)^2\\ =\left(2x\right)^2=4x^2=VP\)