(x+2)^2-(x-3)(x+3)=6

x=-1,75

\(\left(x+2\right)^2-\left(x-3\right)\cdot\left(x+3\right)=6\)

\(x^2+4x+4-x^2+9=6\)

\(4x+13=6\)

\(x=-\frac{7}{4}=-1.75\)

Ta có

\(\frac{1}{x^2+y^2+z^2}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\)

\(=\left(\frac{1}{x^2+y^2+z^2}+\frac{\frac{4}{9}}{2xy}+\frac{\frac{4}{9}}{2yz}+\frac{\frac{4}{9}}{2zx}\right)+\frac{7}{9}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)

\(\ge\frac{\left(1+\frac{2}{3}+\frac{2}{3}+\frac{2}{3}\right)^2}{x^2+y^2+z^2+2xy+2yz+2zx}+\frac{7}{9}.\frac{\left(1+1+1\right)^2}{xy+yz+xz}\)

\(\ge\frac{9}{\left(x+y+z\right)^2}+\frac{7}{9}.\frac{9}{\frac{\left(x+y+z\right)^2}{3}}\)

\(=9+\frac{7}{9}.27=30\)

Vậy GTNN là 30 đạt được khi \(x=y=z=\frac{1}{3}\)

lớp mấy

\(A=5-\left(x^2+2\right)^2-\left(y-1\right)^2\le5\)

A_max =5

A=-x²-4x-4-y²+2y-1+5

A=-(x+2)²-(y-1)²+5

A=-((x+2)²+(y-1)²)+5

Max_A=5

\(\orbr{\begin{cases}y=\frac{3}{x}\\z=\frac{4}{x}\end{cases}\Rightarrow\frac{12}{x^2}=6\Rightarrow x^2=2}\)

\(\orbr{\begin{cases}x=\frac{3}{y}\\z=\frac{6}{y}\end{cases}\Rightarrow\frac{18}{y^2}=4\Rightarrow y^2=\frac{9}{2}}\)

\(\orbr{\begin{cases}x=\frac{4}{z}\\y=\frac{6}{z}\end{cases}\Rightarrow\frac{24}{z^2}=3\Rightarrow z^2=8}\)

\(A=\frac{1}{2}\left(2+\frac{9}{2}+8\right)=\frac{4+9+16}{4}=\frac{29}{4}\) 

\(M=\left(a^2+2ab+b^2-6a-6b+9\right)+\left(b^2-2b+1\right)+2017\)

\(M=\left(a+b-3\right)^2+\left(b-1\right)^2+2017\ge2017\Rightarrow M_{min}=2017\)

ngonhuminh giảng cho minh cách ghep BP khi nhìn đa thức rất lùng tùng với, 

Cái này giống dạng toán thường thôi bạn ạ! Để mình giải cho bạn nhé!

x-2/x-1 + 2x(1-x)/x^2-x = x-2/x-1 + (-2x(x-1))/x(x-1) = x-2/x-1 + (-2x/x) = x^2-2x/x(x-1) +( -2x(x-1)/x(x-1)) = x^2-2x+(-2x(x-1))/x(x-1)

= x^2-2x-2x^2+2x/x(x-1) = -x^2/x(x-1) = -x/x-1 

90 độ bạn nhé

f(1)=1+a+b+c=1

a+b+c=0

f(2)=8+4a+2b+c=4

4a+2b+c=-4 

4a+2b+c-(a+b+c)=-4

3a+b=-4

3(3a+b)=-12

9a+3b=-12 

f(3)=27+9a+3b+c=9

9a+3b+c=-18

-12+c=-18

c=-6

ta lại có 4a+2b+c-4(a+b+c)=-4-4.0=-4

-2b-3c=-4

-2b+18=-4

-2b=-22

b=11

a+b+c=0

a+11-6=0

a+5=0

a=-5

f(x)=x^3-5x^2+11x-6

đến đây bạn tự giải f(6),f(7),f(8) nhan

\(1+a+b+c=1\)(1)

\(8+4a+2b+c=4\)(2)

\(27+9a+3b+c=9\)(3)

a+b+c=0

4a+2b+c=-4

9a+3b+c=-18

---

3a+b=-4

8a+2b=-18

=>2a=-10=> a=5; b=-19;c=14

f(x)=x^2+5x^2-19x+14

f(6)=6^3+5.6^2-19.6+14=

.....

bạn kéo xuống dưới xem bài của bạn Quang Huy Thịnh đi nãy mik vừa giải một bài tương tự như zị 

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\frac{yz}{xyz}+\frac{xz}{xyz}+\frac{xy}{xyz}=0\)

\(\Leftrightarrow\frac{yz+xz+xy}{xyz}=0\Leftrightarrow yz+xz+xy=0\)

\(A=\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}=\frac{xy\left(x+y\right)}{xyz}+\frac{xz\left(x+z\right)}{xyz}+\frac{yz\left(y+z\right)}{xyz}\)

\(=\frac{x^2y+xy^2}{xyz}+\frac{x^2z+xz^2}{xyz}+\frac{y^2z+yz^2}{xyz}=\frac{x^2y+xy^2+x^2z+xz^2+y^2z+yz^2}{xyz}\)

\(=\frac{\left(x^2y+x^2z+xyz\right)+\left(xy^2+y^2z+xyz\right)+\left(xz^2+yz^2+xyz\right)-3xyz}{xyz}\)

\(=\frac{x\left(xy+xz+yz\right)+y\left(xy+yz+xz\right)+z\left(xz+yz+xy\right)-3xyz}{xyz}\)

\(=\frac{\left(x+y+z\right)\left(xz+yz+xy\right)-3xyz}{xyz}=\frac{\left(x+y+z\right).0-3xyz}{xyz}=\frac{-3xyz}{xyz}-3\)