\(\left(a+10\right)+\left(a^2+10a\right)^2+2\left(a+10\right)^2=0\)
Có: \(\left\{{}\begin{matrix}\left(a^2+10a\right)^2\ge0\forall a\\2\left(a+10\right)^2\ge0\forall a\end{matrix}\right.\)
Để bt = 0 => \(\left\{{}\begin{matrix}\left(a^2+10a\right)^2=0\\2\left(a+10\right)^2=0\end{matrix}\right.\)\(\Rightarrow a=-10\)
Thay a = -10 vào a + 10 có: -10 + 10 = 0
(tm)
Vậy a = -10
Rút gọn biểu thức:
a) \(A=\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right).\left(\sqrt{10}-\sqrt{2}\right)\)
b) \(B=\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}+\frac{\sqrt{a}+1}{\sqrt{a}-1}\right).\left(1-\frac{2}{a+1}\right)^2\) với \(a>0,a\ne1\)
ta có \(\sqrt{\left(1+a^3\right)\left(1+b^3\right)}=\sqrt{\left(1+a\right)\left(a^2-a+1\right)}.\sqrt{\left(1+b\right)\left(b^2-b+1\right)}\)
Mà \(\sqrt{\left(a+1\right)\left(a^2-a+1\right)}\le\dfrac{a+1+a^2-a+2}{2}=\dfrac{a^2+2}{2}\)
Tương tự thì \(\sqrt{\left(1+a^3\right)\left(1+b^3\right)}\le\dfrac{\left(a^2+2\right)\left(b^2+2\right)}{4}\Rightarrow\dfrac{a^2}{\sqrt{\left(1+a^3\right)\left(1+B^3\right)}}\ge\dfrac{4a^2}{\left(a^2+2\right)\left(b^2+2\right)}\)
=\(\dfrac{4a^2\left(c^2+2\right)}{\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)}\)
Tương tự rồi + vào, ta có
...\(\ge4\dfrac{a^2\left(c^2+2\right)+b^2\left(a^2+2\right)+c^2\left(b^2+2\right)}{\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)}\)
ta cần chứng minh \(3\left[a^2\left(c^2+2\right)+b^2\left(a^2+2\right)+c^2\left(b^2+2\right)\right]\ge\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)\)
đến đây nhân tung ra và dùng cô-si tiếp
cho a,b,c >0 ; \(a+b+c=a^2+b^2+c^2=2\)
Tinnsh \(A=a\sqrt{\dfrac{\left(1+b^2\right)\left(1+c^2\right)}{1+a^2}}+b\sqrt{\dfrac{\left(1+a^2\right)\left(1+c^2\right)}{1+b^2}}+c\sqrt{\dfrac{\left(1+a^2\right)\left(1+b^2\right)}{1+c^2}}\)
Tính:
a) \(A=\left(\sqrt{6}+\sqrt{10}\right)-\sqrt{4-\sqrt{15}}\)
b) \(B=\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right).\sqrt{3-\sqrt{15}}\)
c) \(C=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4-\sqrt{15}}\)
\(P=\left(\dfrac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
a) Rút gọn P (x > o, x khác 1)
b) Tìm giá trị của x để P > 0
Câu 1 ) Cho \(a,b,c\in R\) . Chứng minh rằng :
M=\(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\dfrac{3\left(a+b+c\right)^2}{4}\)
Câu 2 ) Cho \(a>0;b>0;a+b\le1\) . Tìm GTNN của biểu thức :
A = \(\dfrac{2}{a^2+b^2}+\dfrac{35}{ab}+2ab\)
Câu 3) Cho \(a>0;b>0\) . Chứng minh rằng : \(\left(4a^2+b^2\right)\left(\dfrac{1}{a^2}+\dfrac{1}{4b^2}\right)\ge4\)
Bài 1: Tính
a) \(\sqrt{7-2\sqrt{10}}-\sqrt{6-2\sqrt{5}}\)
b) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
Bài 2
a) Cho x<y<0. Tính (thu gọn) A=\(\sqrt{x^2}+\sqrt{y^2}+\sqrt{\left(x+y\right)^2}\)
b) Tình a,b,c biết a+b+c= \(2\sqrt{a}+2\sqrt{b-3}+2\sqrt{c}\)
Rút gọn:
a) \(A=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{1-\sqrt{x}}{1-x}\right)^2\left(x\ge0,x\ne1\right)\)
b) \(B=\left(\frac{2-a\sqrt{a}}{2-\sqrt{a}}+\sqrt{a}\right)\left(\frac{2-\sqrt{a}}{2-a}\right)\left(a\ge0,a\ne2,a\ne4\right)\)
c) \(C=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\left(x>0,x\ne1\right)\)
Cho \(a^5+ab^4=b^{10}+b^6\)
Tính \(P=\left(a-b^2\right)\left(a+2b^2\right)\)
