Bài 3: Liên hệ giữa phép nhân và phép khai phương

\(a.\sqrt{2a}.\sqrt{18a}=\sqrt{2a}.3\sqrt{2a}=3.2a=6a\)

\(b.\sqrt{3a.27ab^2}=\sqrt{9a^2b^2.9}=9\text{ |}ab\text{ |}\)

\(c.2y^2.\sqrt{\dfrac{x^4}{4y^2}}=2y^2.\dfrac{x^2}{-2y}=-x^2y\)

\(d.\dfrac{y}{x}.\sqrt{\dfrac{x^2}{y^4}}=\dfrac{y}{x}.\dfrac{x}{y^2}=\dfrac{1}{y}\)

\(e.\sqrt{\dfrac{9a^2}{16}}=\dfrac{3\text{ |}a\text{ |}}{4}\)

\(f.\sqrt{10.16a^2}=-4a\sqrt{10}\)

\(g.\sqrt{a^4\left(3-a\right)^2}=a^2\left(a-3\right)\)

\(h.\sqrt{\dfrac{2a^2b^4}{98}}\sqrt{\dfrac{a^2b^4}{49}}=\dfrac{b^2\text{ |}a\text{ |}}{7}\)

ĐKXĐ: x>=0; y>=1 ; z>=2.

câu 1:Từ giả thiết ta có:

\(2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}=x+y+z\)

\(\Leftrightarrow x-2\sqrt{x}+1+\left(y-1\right)-2\sqrt{y-1}+1+\left(z-2\right)-2\sqrt{z-2}+1=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)

\(\Leftrightarrow\sqrt{x}=1;\sqrt{y-1}=1;\sqrt{z-2}=1\)

Vậy x=1;y=2;z=3.

Có gì ko hiểu bạn cứ bình luận phía dưới :)

a)\(pt\Leftrightarrow\sqrt{3x^2-6x+4}+\sqrt{2x^2-4x+6}+x^2-2x-2=0\)

\(\Leftrightarrow\sqrt{3x^2-6x+4}-1+\sqrt{2x^2-4x+6}-2+x^2-2x+1=0\)

\(\Leftrightarrow\dfrac{3x^2-6x+4-1}{\sqrt{3x^2-6x+4}+1}+\dfrac{2x^2-4x+6-4}{\sqrt{2x^2-4x+6}+2}+\left(x-1\right)^2=0\)

\(\Leftrightarrow\dfrac{3\left(x-1\right)^2}{\sqrt{3x^2-6x+4}+1}+\dfrac{2\left(x-1\right)^2}{\sqrt{2x^2-4x+6}+2}+\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(\dfrac{3}{\sqrt{3x^2-6x+4}+1}+\dfrac{2}{\sqrt{2x^2-4x+6}-2}+1\right)=0\)

Dễ thấy: \(\dfrac{3}{\sqrt{3x^2-6x+4}+1}+\dfrac{2}{\sqrt{2x^2-4x+6}-2}+1>0\)

\(\Rightarrow\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)

b)\(\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+9}=3-4x-2x^2\)

\(pt\Leftrightarrow\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+9}+2x^2+4x-3=0\)

\(\Leftrightarrow\sqrt{3x^2+6x+12}-3+\sqrt{5x^4-10x^2+9}-2+2x^2+4x-8=0\)

\(\Leftrightarrow\sqrt{3x^2+6x+12}-3+\sqrt{5x^4-10x^2+9}-2+2x^2+4x+2=0\)

\(\Leftrightarrow\dfrac{3x^2+6x+12-9}{\sqrt{3x^2+6x+12}+3}+\dfrac{5x^4-10x^2+9-4}{\sqrt{5x^4-10x^2+9}+2}+2\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow\dfrac{3\left(x+1\right)^2}{\sqrt{3x^2+6x+12}+3}+\dfrac{5\left(x+1\right)^2\left(x-1\right)^2}{\sqrt{5x^4-10x^2+9}+2}+2\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(x+1\right)^2\left(\dfrac{3}{\sqrt{3x^2+6x+12}+3}+\dfrac{5\left(x-1\right)^2}{\sqrt{5x^4-10x^2+9}+2}+2\right)=0\)

Dễ thấy: \(\dfrac{3}{\sqrt{3x^2+6x+12}+3}+\dfrac{5\left(x-1\right)^2}{\sqrt{5x^4-10x^2+9}+2}+2>0\)

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

\(a.11+2\sqrt{30}=6+2.\sqrt{6}.\sqrt{5}+5=\left(\sqrt{6}+\sqrt{5}\right)^2\)

\(b.10+2\sqrt{21}=7+2\sqrt{7}.\sqrt{3}+3=\left(\sqrt{7}+\sqrt{3}\right)^2\)

\(c.6x-\sqrt{x}-1=6x+2\sqrt{x}-3\sqrt{x}-1=2\sqrt{x}\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}+1\right)=\left(3\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\) \(d.14-2\sqrt{45}=14-6\sqrt{5}=9-2.3\sqrt{5}+5=\left(3-\sqrt{5}\right)^2\)

\(e.4x-3\sqrt{x}-1=4x-4\sqrt{x}+\sqrt{x}-1=4\sqrt{x}\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)=\left(\sqrt{x}-1\right)\left(4\sqrt{x}+1\right)\) \(j.12-2\sqrt{27}=9-2.3.\sqrt{3}+3=\left(3-\sqrt{3}\right)^2\)

a)

\(M=\sqrt{3\sqrt{2}+2\sqrt{3}}\times\sqrt{3\sqrt{2}-2\sqrt{3}}\)

\(M^2=\sqrt{2\times3}\left(\sqrt{3}+\sqrt{2}\right)\times\sqrt{2\times3}\left(\sqrt{3}-\sqrt{2}\right)\)

\(=6\)

\(\Rightarrow M=\sqrt{6}\)

b)

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

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

\(=1-3+2\sqrt{6}-2=-4+2\sqrt{6}\)

c)

\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\)

\(=\left(5+4\sqrt{2}\right)\left(9-4-4\sqrt{2}\right)\)

\(=25-32=-7\)

Ta có:C= \(\dfrac{\sqrt{x}}{\sqrt{x}+1}\) =\(\dfrac{\sqrt{x}+1-1}{\sqrt{x}+1}\) =1-\(\dfrac{1}{\sqrt{x}+1}\)

Để C min <=> 1-\(\dfrac{1}{\sqrt{x}+1}\) min

<=> \(\dfrac{1}{\sqrt{x}+1}\) max

<=> \(\sqrt{x}\) +1 min

Dấu''='' xảy ra <=> \(\sqrt{x}\) +1=1

<=> \(\sqrt{x}\) =0

<=> x=0

Vậy C min=0 <=> x=0

Ta có: P=\(\dfrac{\sqrt{x}+4}{\sqrt{x}+1}\) = \(\dfrac{\sqrt{x}+1+3}{\sqrt{x}+1}\)= 1 + \(\dfrac{3}{\sqrt{x}+1}\)

Để P nguyên <=> \(\dfrac{3}{\sqrt{x}+1}\) nguyên

<=> 3 chia hết cho \(\sqrt{x}\) +1

<=> \(\sqrt{x}\)+1 thuộc Ư(3)

Ta có bảng sau:

Vậy S={ 0;4}

\(A=x-\sqrt{x-2018}=x-2018-\sqrt{x-2018}+2018=\left(\sqrt{x-2018}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}+2018\ge2018-\dfrac{1}{4}\)

d=xrk.....

đến đó hiểu rồi chứ? sorry, tại bây giờ máy tính tay 0 nằm trong tầm tay của tớ. Lười đi lấy