a + 2b = 1 => 2b = 1 - a

Biến đổi VT:

\(a^3+8b^3+2ab-a^2-4b^2\)

\(=a\left(a^2+2b-a\right)+\left(2b\right)^3-\left(2b\right)^2\)

\(=a\left(a^2+1-a-a\right)+\left(2b\right)^2\left(2b-1\right)\)

\(=a\left(a^2-2a+1\right)+\left(1-a\right)^2\left(1-a-1\right)\)

\(=a\left(a-1\right)^2-a\left(1-a\right)^2\)

\(=a\left[\left(a-1\right)^2-\left(1-a\right)^2\right]\)

\(=a\left(a-1+1-a\right)\left(a-1-1+a\right)\)

\(=0\)(đpcm)

\(A=\frac{2ab}{4ab}+\frac{2ab}{a^2+4b^2}+\frac{1}{8ab}-\frac{1}{2}\)

áp dụng bđt AM-GM , a,b> 0

\(\Rightarrow A\ge2ab\left(\frac{4}{4ab+a^2+4b^2}\right)+\frac{1}{8ab}-\frac{1}{2}\)

\(\Rightarrow A\ge\frac{8ab}{1}+\frac{1}{8ab}-\frac{1}{2}\)

\(\Rightarrow A\ge2-\frac{1}{2}=\frac{3}{2}\)

Đặt \(\left(\dfrac{1}{a};\dfrac{1}{2b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=0\)

\(M=\dfrac{x^2}{yz}+\dfrac{y^2}{zx}+\dfrac{z^2}{xy}=\dfrac{x^3+y^3+z^3}{xyz}\)

\(=\dfrac{\left(x+y\right)^3-3xy\left(x+y\right)+z^3}{xyz}=\dfrac{-z^3-3xy\left(-z\right)+z^3}{xyz}\)

\(=\dfrac{3xyz}{xyz}=3\)

Điện​thọi bé tý khi viết lời giải chẳng thẫy đề đâu. Vp (a+b)^3=bó tay

=1 phải ko?

Câu 1:

\(Q=a^2+4b^2-10a\)

\(=a^2-10a+25+4b^2-25\)

\(=\left(a-5\right)^2+4b^2-25\)

\(\left(a-5\right)^2\ge0\)

\(4b^2\ge0\)

\(\Rightarrow\left(a-5\right)^2+4b^2-25\ge-25\)

Dấu ''='' xảy ra khi \(\left[\begin{array}{nghiempt}a-5=0\\b=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}a=5\\b=0\end{array}\right.\)

\(MinQ=-25\Leftrightarrow a=5;b=0\)

Câu 2:

Tam giác DAC vuông tại D có:

\(AC^2=CD^2+AD^2\)

\(=CD^2+CD^2\) (ABCD là hình vuông)

\(=2CD^2\)

\(=2\times\left(3\sqrt{2}\right)^2\)

\(=2\times9\times2\)

\(=36\)

\(AC=\sqrt{36}=6\left(cm\right)\)

Câu 3:

\(\frac{1}{a-1}=1\)

\(a-1=1\)

\(a=1+1\)

\(a=2\)

Thay a = 2 vào P, ta có:

\(P=\frac{2-2\times2\times b-b}{2\times2+3\times2\times b-b}\)

\(=\frac{2-4b-b}{4+6b-b}\)

\(=\frac{2-5b}{4+5b}\)