Từ a và b bạn tích ra các số còn lai rồi nhân lại bằng máy tính là được mà bạn^^

a)\(\Leftrightarrow\)\(7\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=8\)

 \(\Leftrightarrow\) \(3\sqrt{x-2}=8\)

  \(\Leftrightarrow\) \(\sqrt{x-2}=24\)

\(\Leftrightarrow\)\(x-2=576\)\(\Leftrightarrow x=578\)

c)\(\Leftrightarrow GTTĐ\left(x-1\right)=\sqrt{2}-1\)\(TH1:x-1>0\)

\(\Rightarrow x-1=\sqrt{2}-1\)\(\Leftrightarrow x=\sqrt{2}\)

\(TH2:x-1< 0\)

\(\Rightarrow1-x=\sqrt{2}-1\)

\(\Leftrightarrow x=2+\sqrt{2}\)

d)\(TH1:x-10=0\Rightarrow x=10\)

\(TH2:\sqrt{x-4}=0\Rightarrow x=4\)

câu b) thì mik cần thêm time

Đặt \(A=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)

\(\Rightarrow A^2=\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2\)

\(=2+\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}2-\sqrt{3}\)

\(=4-2\sqrt{1}\)

\(=2\)

\(\Rightarrow A=\sqrt{2}\)( VÌ \(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}>0\))

Bài này cs nhiều cách lm nha bn!!

Đặt : \(A=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)

\(\sqrt{2}A=\sqrt{2}\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)\)

\(=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)

\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=|\sqrt{3}+1|-|\sqrt{3}-1|\)

\(=\sqrt{3}+1-\sqrt{3}+1\)

\(=2\)

\(\Rightarrow A=\frac{2}{\sqrt{2}}=\sqrt{2}\)

=.= hok tốt!!

a/ \(B=\frac{1+x}{1+\sqrt{x}+x}\)

b/ Giải phương trình bậc 2 thì dễ rồi ha

c/ \(\frac{1+x}{1+\sqrt{x}+x}>\frac{2}{3}\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2>0\)đung vì x khac 1

Phương trình bậc hai là\(x-\sqrt{6x}+1=0\) thì giải làm sao bạn ơi??

Xửa đề thành tìm nghiệm nguyên rồi làm

\(x^2+xy-2008x-2009y-2010=0\)

\(\Leftrightarrow\left(x-2009\right)\left(x+y+1\right)=1\)

làm nôt

a) Ta có:

\(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)

\(=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)

\(=a^2b^2+b^2d^2+a^2d^2+b^2c^2\)

\(=\left(a^2+b^2\right)\left(c^2+d^2\right)\)

b) theo a) \(\Rightarrow\)\(\left(a^2+b^2\right)\left(c^2+d^2\right)\ge\left(ac+bd\right)^2\)

Dấu bằng xảy ra khi ad=bc => a/b=c/d

a,\(\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)

\(=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)=\left(c^2+d^2\right)\left(a^2+b^2\right)\)

b,Xét hiệu

\(\left(a^2+b^2\right)\left(c^2+d^2\right)-\left(ac+bd\right)^2=\left(ad-bc\right)^2\ge0\)

\(\Rightarrow\left(ac+bd\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)\)

Đặt \(2018=a\)

\(\Rightarrow\sqrt{1+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}=\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)

\(=\sqrt{\frac{\left(a^2+a+1\right)^2}{\left(a+1\right)^2}}+\frac{a}{a+1}=\frac{a^2+a+1}{a+1}+\frac{a}{a+1}=\frac{\left(a+1\right)^2}{a+1}=a+1=2019\)

\(\sqrt{\frac{1+2x\sqrt{1-x^2}}{2}}=1-2x^2\)

\(\Leftrightarrow\sqrt{\frac{x^2+2x\sqrt{1-x^2}+1-x^2}{2}}=1-2x^2\)

\(\Leftrightarrow\sqrt{\frac{\left(x+\sqrt{1-x^2}\right)^2}{2}}=1-2x^2\)

\(\Leftrightarrow\frac{x+\sqrt{1-x^2}}{\sqrt{2}}=1-2x^2\)

Làm nôt