Ôn tập cuối năm phần số học

\(x^2-2xy+y^2+1=\left(x^2-2xy+y^2\right)+1=\left(x-y\right)^2+1>0\) nhé!

\(x-x^2-1=-\left(x^2-x+1\right)=-\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}\right)-\dfrac{3}{4}=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}< 0\)

giải bất phương trình

a: =>-4x>16

=>x<-4

c: =>20x-25<=21-3x

=>23x<=46

=>x<=2

d: =>20(2x-5)-30(3x-1)<12(3-x)-15(2x-1)

=>40x-100-90x+30<36-12x-30x+15

=>-50x-70<-42x+51

=>-8x<121

=>x>-121/8

a) \(A=\dfrac{-45.58-45.42}{2+4+6+...+16+18}\)

\(A=-\dfrac{45\left(58+42\right)}{\dfrac{\left(18-2\right):2+1}{2}.\left(2+18\right)}\)

\(A=\dfrac{-45.100}{\dfrac{5}{2}.20}\)

\(A=-\dfrac{45.100}{50}\)

\(A=-90\)

b) \(1-\left[\left(5\dfrac{3}{8}+x-7\dfrac{5}{24}\right):\left(-16\dfrac{2}{3}\right)\right]=0\)

\(\Rightarrow\left(5\dfrac{3}{8}+x-7\dfrac{5}{24}\right):\left(-16\dfrac{2}{3}\right)=1\)

\(\Rightarrow\left(\dfrac{43}{8}+x-\dfrac{173}{24}\right):\left(-\dfrac{46}{3}\right)=1\)

\(\Rightarrow\left(\dfrac{129}{24}-\dfrac{173}{24}+x\right).\left(-\dfrac{3}{46}\right)=1\)

\(\Rightarrow\left(-\dfrac{44}{24}+x\right).\left(-\dfrac{3}{46}\right)=1\)

\(\Rightarrow\left(-\dfrac{11}{6}+x\right).\left(-\dfrac{3}{46}\right)=1\)

\(\Rightarrow-\dfrac{11}{6}+x=1.\left(-\dfrac{46}{3}\right)\)

\(\Rightarrow-\dfrac{11}{6}+x=-\dfrac{46}{3}\)

\(\Rightarrow x=-\dfrac{46}{3}+\dfrac{11}{6}\)

\(\Rightarrow x=-\dfrac{92}{6}+\dfrac{11}{6}\)

\(\Rightarrow x=-\dfrac{81}{6}\)

\(\Rightarrow x=-\dfrac{27}{2}\)

\(x+y-2\left(\sqrt{x}+\sqrt{y}\right)+2\ge0\)

\(\Leftrightarrow x+y-2\sqrt{x}-2\sqrt{y}+2\ge0\)

\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-2\sqrt{y}+1\right)\ge0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y}-1\right)^2\ge0\)

Do : \(\left\{{}\begin{matrix}\left(\sqrt{x}-1\right)^2\ge0\\\left(\sqrt{y}-1\right)^2\ge0\end{matrix}\right.\Rightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y}-1\right)^2\ge0\)

Vậy đẳng thức được chứng minh !

Câu 3: 

a: \(B=\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\)

\(=\dfrac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^3-x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)

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

b: Để B=0 thì x-1=0

hay x=1

c: Để B>1 thì x-1>1

=>x>2

+) ta có : \(E=3x^2-6x+15=3\left(x^2-2x+1\right)+12\)

\(=3\left(x-1\right)^2+12\ge12\) \(\Rightarrow E_{min}=12\) khi \(x=1\)

+) ta có : \(F=5x^2+6x-12=5\left(x^2+\dfrac{6}{5}x+\dfrac{9}{25}\right)-\dfrac{69}{5}\)

\(=5\left(x+\dfrac{3}{5}\right)^2-\dfrac{69}{5}\ge\dfrac{-69}{5}\) \(\Rightarrow F_{min}=-\dfrac{69}{5}\) khi \(x=\dfrac{-3}{5}\)

+) ta có : \(G=4x^2-4x+25=4\left(x^2-x+\dfrac{1}{4}\right)+24\)

\(=4\left(x-\dfrac{1}{2}\right)^2+24\ge24\) \(\Rightarrow G_{min}=24\) khi \(x=\dfrac{1}{2}\)

+) ta có : \(H=9x^2+6x^2+4=15x^2+4\ge4\)

\(\Rightarrow H_{min}=4\) khi \(x=0\)

Tìm GTNN

E=3x^2-6x+15

F= 5x^2+6x-12

G=4x^2-4x+25

H=9x^2+6x^2+4

Lời giải:

Ta có: \(4x^2-4xy+y^2=0\)

\(\Leftrightarrow (2x)^2-2.2x.y+y^2=0\)

\(\Leftrightarrow (2x-y)^2=0\Rightarrow 2x=y\)

Thay vào biểu thức $P$ ta thu được:

\(P=\frac{x+y}{x-y}=\frac{x+2x}{x-2x}=\frac{3x}{-x}=-3\)

Bài 1:

a) \(A=\left[-2\left(m+1\right)\right]^2-4\left(m-4\right)\)

\(A=\left(-2\right)^2\left(m+1\right)^2-4m+16\)

\(A=4\left(m^2+2m+1\right)-4m+16\)

\(A=4m^2+8m+4-4m+16\)

\(A=4m^2+4m+20\)

\(A=\left(2m\right)^2+2.m.2+4+16\)

\(A=\left(2m+2\right)^2+16\)

Vì \(\left(2m+2\right)^2\ge0\) với mọi m

\(\Rightarrow\left(2m+2\right)^2+16\ge16\)

Mà \(16>0\)

\(\Rightarrow\left(2m+2\right)^2+16>0\)

\(\Rightarrow\left[-2\left(m+1\right)\right]^2-4\left(m-4\right)>0\) ( Đpcm )

b) \(B=\left[-\left(m+2\right)\right]^2-4.2m\)

\(B=\left(-1\right)^2\left(m+2\right)^2-8m\)

\(B=m^2+2m.2+4-8m\)

\(B=m^2+4m+4-8m\)

\(B=m^2-4m+4\)

\(B=m^2-2.m.2+2^2\)

\(B=\left(m-2\right)^2\)

Vì \(\left(m-2\right)^2\ge0\) với mọi m

\(\Rightarrow\left[-\left(m+2\right)\right]^2-4.2m\ge0\) ( Đpcm )

c) \(C=\left(m+1\right)^2-4.2.\left[-\left(m+3\right)\right]\)

\(C=m^2+2m+1-8\left(-m-3\right)\)

\(C=m^2+2m+1+8m+24\)

\(C=m^2+10m+25\)

\(C=\left(m+5\right)^2\)

Vì \(\left(m+5\right)^2\ge0\) với mọi m

\(\Rightarrow\left(m+1\right)^2-4.2\left[-\left(m+3\right)\right]\ge0\) ( Đpcm )