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

a)\(\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right)\div\sqrt{15}=\left(2\sqrt{3}+5\sqrt{3}+3\sqrt{3}\right)\div\sqrt{3}\sqrt{5}=10\sqrt{3}\div\sqrt{3}\sqrt{5}=\sqrt{2}\sqrt{5}\div\sqrt{5}=\sqrt{2}\)b)\(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}=\sqrt{4}\sqrt{9}\sqrt{7}-\sqrt{100}\sqrt{7}+\sqrt{16}\sqrt{9}\sqrt{7}-\sqrt{64}\sqrt{7}=2\cdot3\cdot\sqrt{7}-10\cdot\sqrt{7}+4\cdot3\cdot\sqrt{7}-8\sqrt{7}=6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}=0\)

c)\(\sqrt{27^2-23^2}+\sqrt{37^2-35^2}=\sqrt{\left(27-23\right)\left(27+23\right)}+\sqrt{\left(37-35\right)\left(37+35\right)}=\sqrt{4\cdot50}\cdot\sqrt{2\cdot72}=\sqrt{4\cdot50\cdot2\cdot72}=\sqrt{2^2\cdot2\cdot25\cdot2\cdot36\cdot2}=\sqrt{16}\cdot\sqrt{25}\cdot\sqrt{36}=4\cdot5\cdot6=120\)

d)\(\left(\sqrt{\dfrac{1}{7}}+\sqrt{\dfrac{16}{7}}+\sqrt{\dfrac{9}{7}}\right)\div\sqrt{7}=\left(\dfrac{1}{\sqrt{7}}+\dfrac{4}{\sqrt{7}}+\dfrac{3}{\sqrt{7}}\right)\cdot\dfrac{1}{\sqrt{7}}=\dfrac{7}{\sqrt{7}}\cdot\dfrac{1}{\sqrt{7}}=1\)

\(A=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3x^2+6xy+3y^2}{4}}=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3\left(x^2++2xy+y^2\right)}{4}}=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3\left(x-y\right)^2}{4}}=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\sqrt{3}\left(x-y\right)}{2}=\dfrac{\sqrt{3}}{x+y}\)

\(B=\dfrac{1}{2a-1}\cdot\sqrt{5a^4\left(1-4a+4a^2\right)}=\dfrac{1}{2a-1}\cdot\sqrt{5a^4\left(2a-1\right)^2}=\dfrac{1}{2a-1}\cdot\sqrt{5}a^2\left(2a-1\right)=\sqrt{5}\cdot a^2\)

đề là j nhỉ bạn

\(\dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)

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

\(=\dfrac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{\sqrt{5-1}}-\sqrt{2-2\sqrt{2}+1}\)

Đặt: \(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}\)

\(\Rightarrow A^2=3+\sqrt{5}+2\sqrt{\left(3+\sqrt{5}\right)\left(7-3\sqrt{5}\right)}+7-3\sqrt{5}\)

\(A^2=10-2\sqrt{5}+2\sqrt{6-2\sqrt{5}}\)

\(A^2=10-2\sqrt{5}+2\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}+1}\)

\(A^2=10-2\sqrt{5}+2\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(A^2=10-2\sqrt{5}+2\left|\sqrt{5}-1\right|\)

\(A^2=10-2\sqrt{5}+2\sqrt{5}-2\)

\(A^2=8\Rightarrow A=2\sqrt{2}\)

Thay vào ta có:

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

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

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

√√5+2+√√5−2√√5+1−√3−2√2

=(√√5+2+√√5−2)(√√5−1)(√√5+1)(√√5−1)−√3−2√2

=√3+√5+√7−3√5√5−1−√2−2√2+1

Đặt: A=√3+√5+√7−3√5

⇒A2=3+√5+2√(3+√5)(7−3√5)+7−3√5

A2=10−2√5+2√6−2√5

A2=10−2√5+2√(√5)2−2√5+1

A2=10−2√5+2√(√5−1)2

A2=10−2√5+2|√5−1|

A2=10−2√5+2√5−2

A2=8⇒A=2√2

Thay vào ta có:

2√22−√(√2−1)2

=√2−|√2−1|

=√2−(√2−1)=√2−√2+1=1

viết phần k đc

a) điều kiện \(x\ge0;x\ne4;x\ne9\)

\(Q=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

\(Q=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)

\(Q=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(Q=\dfrac{2\sqrt{x}-9-\left(x-9\right)+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(Q=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(Q=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\Leftrightarrow\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(Q=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

b) ta có : \(Q=0\) \(\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=0\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}+1=0\\\sqrt{x}-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=-1\\\sqrt{x}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\in\varnothing\\x=9\left(loại\right)\end{matrix}\right.\) vậy không có giá trị nào của x để Q = 0

c) ta có : \(x=\sqrt{7+\sqrt{24}}\Leftrightarrow x=\sqrt{\left(\sqrt{6}+1\right)^2}\Leftrightarrow x=\sqrt{6}+1\)

\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{6}+1\right)}\)

thay vào Q ta có \(Q=\dfrac{\sqrt{\left(\sqrt{6}+1\right)}+1}{\sqrt{\left(\sqrt{6}+1\right)}-3}\)

d) ta có : \(Q>0\) \(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}>0\)

mà ta có : \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1>0\)

\(\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}>0\Leftrightarrow\sqrt{x}-3>0\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)

vậy \(x\ge9\) thì \(Q>0\)

Áp dụng BĐT Cauchy cho các số không âm , ta có :

\(\dfrac{a}{b}+\dfrac{b}{c}\) ≥ \(2\sqrt{\dfrac{a}{b}.\dfrac{b}{c}}=2\sqrt{\dfrac{a}{c}}\left(1\right)\)

\(\dfrac{b}{c}+\dfrac{c}{a}\) ≥ \(2\sqrt{\dfrac{b}{c}.\dfrac{c}{a}}=2\sqrt{\dfrac{b}{a}}\left(2\right)\)

\(\dfrac{a}{b}+\dfrac{c}{a}\) ≥ \(2\sqrt{\dfrac{a}{b}.\dfrac{c}{a}}=2\sqrt{\dfrac{c}{b}}\left(3\right)\)

Cộng từng vế của ( 1 ; 2 ; 3) , ta có :

\(2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\) ≥ \(2\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)\)

⇔ \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\) ≥ \(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\)

a, DK X>2

x-2=16

x=18

b,DK X>5

ma -3 <0

suy ra vo nghiem

câu a là j có b mà điều kiện b < 2

b: \(=\left|b\cdot\left(b-1\right)\right|=b\cdot\left|b-1\right|\)

c: \(=\left|a\right|\cdot\left|a+1\right|=a\left(a+1\right)=a^2+a\)

d: \(=1-2a-4a=-6a+1\)