c) \(C=4-x^2+2x\)

\(\Leftrightarrow C=-\left(x^2-2x+4\right)+8\)

\(\Leftrightarrow C=-\left(x-2\right)^2+8\le8\)

Max \(C=8\Leftrightarrow x=2\)

Mấy câu kia tương tự ,bạn làm nhé

a\(A=x^2-3x+5\)

\(\Leftrightarrow A=x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}+5-\dfrac{9}{4}\)

\(\Leftrightarrow A=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)

Min \(A=\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{2}\)

1.You can turn off the TV . I am not watch it.

2. What did you do last night ? Well, as soon as I finished my work , I went to bed.

Bài 2: a) \(\left(\dfrac{-1}{2}\right)^2.\dfrac{1}{4}-2.\left(\dfrac{-1}{2}\right)^2\)

\(\Leftrightarrow\left(\dfrac{-1}{2}\right)^2.\left(\dfrac{1}{4}-2\right)\)

\(=\dfrac{1}{4}.\dfrac{-7}{4}=\dfrac{-7}{16}\)

Mình bận rồi , chút làm nhé

a) \(64< 2^n< 256\)

\(\Leftrightarrow2^6< 2^n< 2^8\)

\(\Rightarrow n=7\)

b) \(32\ge2^n>1\)

\(\Leftrightarrow2^5\ge2^n>1\)

\(\Rightarrow n=\left\{1;2;3;4;5\right\}\)

c,d) tương tự

\(B=\left(x-2\right).\left(x-5\right).\left(x^2-7x-10\right)\)

\(\Leftrightarrow B=\left(x^2-7x+10\right).\left(x^2-7x-10\right)\)

\(\Leftrightarrow B=\left(x^2-7x\right)^2-100\ge-100\)

Min B= -100 <=> x=0

Bài 3: \(A=x-x^2\)

\(\Leftrightarrow A=-x^2+2.\dfrac{1}{2}x-\dfrac{1}{4}+\dfrac{1}{4}\)

\(\Leftrightarrow A=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

Max \(A=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)

b) \(N=2x-2x^2-5\)

\(\Leftrightarrow N=-2x^2+2x-5\)

\(\Leftrightarrow N=-2.\left(x^2-x+\dfrac{5}{2}\right)\)

\(\Leftrightarrow N=-2.\left(x^2-2.\dfrac{1}{2}.x+\dfrac{1}{4}+\dfrac{9}{4}\right)\)

\(\Leftrightarrow N=-2.\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le\dfrac{-9}{2}\)

Max \(N=\dfrac{-9}{2}\Leftrightarrow x=\dfrac{1}{2}\)

Câu 1)\(H=\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)

\(\Leftrightarrow H=\left(x-y+z+z-y\right)^2\)

\(\Leftrightarrow H=\left(x-2y+2z\right)^2\)

Câu 2: \(Q=2x^2-6x\)

\(\Leftrightarrow Q=2\left(x^2-2.\dfrac{3}{2}.x+\left(\dfrac{3}{2}\right)^2\right)-\dfrac{9}{2}\)

\(\Leftrightarrow Q=2.\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge\dfrac{-9}{2}\)

Min \(Q=\dfrac{-9}{2}\Leftrightarrow x=\dfrac{3}{2}\)

Ta có: x-y=4 => x=4+y (1)

xy=5 => x=5/y (2)

Từ 1 và 2 => \(\dfrac{5}{y}=4+y\)

\(\Leftrightarrow\dfrac{5}{y}=\dfrac{\left(4+y\right).y}{y}\)

\(\Rightarrow y^2+4y-5=0\)

\(\Leftrightarrow y^2+5y-y-5=0\)

\(\Leftrightarrow\left(y+5\right).\left(y-1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}y=-5\Rightarrow x=-1\\y=1\Rightarrow x=5\end{matrix}\right.\)

Mà x,y <0 nên ta lấy \(x=-1,y=-5\)

=> \(x+y=-1+-5=-6\)