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\(x^2+y^2-xy=x+y-1\)

\(\Leftrightarrow2\left(x^2+y^2-xy\right)=2\left(x+y-1\right)\)

\(\Leftrightarrow2x^2+2y^2-2xy=2x+2y-2\)

\(\Leftrightarrow2x^2+2y^2-2xy-2x-2y+2=0\)

\(\Leftrightarrow x^2+x^2+y^2+y^2-2xy-2x-2y+1+1=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2-2y+1\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(x-1\right)^2+\left(y-1\right)^2=0\)

Vì mũ chẵn luôn lớn hơn hoặc bằng 0

\(\Rightarrow\hept{\begin{cases}x-y=0\\x-1=0\\y-1=0\end{cases}\Rightarrow\hept{\begin{cases}1-1=0\left(tm\right)\\x=1\\y=1\end{cases}}}\)

\(\Rightarrow M=\left(1+1\right)^{10}=2^{10}=1024\)

A B C H D E M K

a) Qua A kẻ đường thẳng vuông góc với tia DE tại K.

Xét tứ giác AHDK: ^AHD = ^HDK = ^AKD = 90⁰; AH=DH => AHDK là hình vuông

=> ^HAK = 90⁰ và AH=AK

Ta có: ^BAH + ^HAC = ^EAK + ^HAC = 90⁰ => ^BAH = ^EAK

Xét \(\Delta\)AHB và \(\Delta\)AKE có: ^AHB = ^AKE (=90⁰); AH=AK; ^BAH = ^EAK

=> \(\Delta\)AHB = \(\Delta\)AKE (g.c.g) => AB=AE (2 cạnh tương ứng) (đpcm).

b) Xét \(\Delta\)ABE vuông tại A có trung tuyến AM => AM=BE/2. Tương tự: DM=BE/2

=> AM=DM => \(\Delta\)MAH = \(\Delta\)MDH (c.c.c) => ^AHM = ^DHM = ^AHD/2 = 45⁰.

ĐS...

Bạn tự xét dấu "=" nhé, mình chỉ hướng dẫn cách tách thôi

a) \(A=5x^2-4x+1\)

\(A=5\left(x^2-\frac{4}{5}x+\frac{1}{5}\right)\)

\(A=5\left[x^2-2\cdot x\cdot\frac{2}{5}+\left(\frac{2}{5}\right)^2+\frac{1}{25}\right]\)

\(A=5\left[\left(x-\frac{2}{5}\right)^2+\frac{1}{25}\right]\)

\(A=5\left(x-\frac{2}{5}\right)^2+\frac{1}{5}\ge\frac{1}{5}\forall x\)

b) Tương tự đặt -9 ra ngoài rồi khai triển như câu a)

c) \(F=-2x^2-y^2+2xy+4x-40\)

\(F=-x^2-x^2-y^2+2xy+4x-40\)

\(F=-\left(x^2-2xy+y^2\right)-\left(x^2-4x+4\right)-36\)

\(F=-36-\left(x-y\right)^2-\left(x-2\right)^2\)

\(F=-36-\left[\left(x-y\right)^2+\left(x-2\right)^2\right]\le-36\forall x;y\)

a+b+c=1 => (a+b+c)²=1

=>a²+b²+c²+2(ab+bc+ca)=1

=>1+2(ab+bc+ca)=1

=>ab+bc+ca=0

Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\Rightarrow x=ak,y=bk,z=ck\)

\(A=xy+yz+zx=akbk+bkck+ckak=k^2\left(ab+bc+ca\right)=0\)

\(A+3=\left(1+\frac{x+y}{z}\right)+\left(1+\frac{x+z}{y}\right)+\left(1+\frac{y+z}{x}\right)\)

\(A+3=\left(x+y+z\right).\left(\frac{1}{z}+\frac{1}{y}+\frac{1}{x}\right)\)

\(A+3=\left(x+y+z\right).0=0\Rightarrow A=-3\)

\(A=\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}=\left(\frac{x+y}{z}+1\right)+\left(\frac{x+z}{y}+1\right)+\left(\frac{y+z}{x}+1\right)-3\)

\(=\frac{x+y+z}{z}\cdot\frac{x+y+z}{y}\cdot\frac{x+y+z}{x}-3=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-3=-3\)

Có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{3}\)

\(\Rightarrow\frac{xy+yz+zx}{xyz}=\frac{1}{3}\)

\(\Rightarrow3.\left(xy+yz+zx\right)=xyz\)(1)

Lại có: \(x+y+z=3\)

\(\Rightarrow\left(x+y+z\right)^2=3^2\)

\(\Rightarrow x^2+y^2+z^2+2xy+2yz+2zx=9\)

Mà: \(x^2+y^2+z^2=17\)

\(\Rightarrow17+2xy+2yz+2xz=9\)

\(\Rightarrow2xy+2yz+2xz=-8\)

\(\Rightarrow xy+yz+zx=-4\)(2)

Thay (2) vào (1) ta có:

\(3.\left(-4\right)=xyz\)

\(xyz=-12\)

Vậy \(xyz=-12\)

Tham khảo nhé~

\(M=x^2-2x+2014\)

\(M=x^2-2\cdot x\cdot1+1^2+2013\)

\(M=\left(x-1\right)^2+2013\ge2013\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy M_min = 2013 khi và chỉ khi x = 1