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

Lời giải:

ĐK: \(x\geq \frac{-1}{4}\)

Biến đổi:

\(x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}=\left(x+\frac{1}{4}\right)+2.\frac{1}{2}\sqrt{x+\frac{1}{4}}+(\frac{1}{2})^2\)

\(=(\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2\)

\(\Rightarrow \sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=\sqrt{x+\frac{1}{4}}+\frac{1}{2}\)

Do đó pt ban đầu tương đương với:

\(x+\sqrt{x+\frac{1}{4}}+\frac{1}{2}=2\)

\(\Leftrightarrow (x+\frac{1}{4})+\sqrt{x+\frac{1}{4}}+\frac{1}{4}=2\)

\(\Leftrightarrow (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2=2\)

\(\Rightarrow \sqrt{x+\frac{1}{4}}+\frac{1}{2}=\sqrt{2}\) (TH bằng $-\sqrt{2}$ ta có thể loại luôn vì biểu thức không âm)

\(\Rightarrow x+\frac{1}{4}=(\sqrt{2}-\frac{1}{2})^2=\frac{9-4\sqrt{2}}{4}\)

\(\Rightarrow x=2-\sqrt{2}\) (t/m)

Vậy..............

@Phùng Khánh Linh Cậu ơi giúp tớ với.

A = \(\sqrt{2}\left(\sqrt{8}-\sqrt{32}-2\sqrt{18}\right)=\sqrt{16}-\sqrt{64}-2\sqrt{36}=4-8-2\cdot6=-4-12=-16\)

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\(B=\sqrt{2}-\sqrt{3-\sqrt{5}}=\dfrac{2-\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\dfrac{2-\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\dfrac{2-\sqrt{5}+1}{\sqrt{2}}=\dfrac{3-\sqrt{5}}{\sqrt{2}}\)

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\(C=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\dfrac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}-\dfrac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\)

còn lại lúc nx mk lm nốt nhé, h bận

\(\sqrt{4+2\sqrt{3}}-\sqrt{5+2\sqrt{6}}+\sqrt{2}\)

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

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

= 1

Vậy biểu thức đã cho bằng 1

\(\left(\sqrt{5-2\sqrt{5}}+\sqrt{2}\right)\sqrt{3}=\sqrt{15-6\sqrt{5}}+\sqrt{6}=\sqrt{5-2.3.\sqrt{5}+9+1}+\sqrt{6}\)

\(=\sqrt{\left(\sqrt{5}-3\right)^2+1}+\sqrt{6}\)

\(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}=\sqrt{49-2.7.3\sqrt{5}+45}-\sqrt{49+2.7.3\sqrt{5}+45}=7-3\sqrt{5}-7-3\sqrt{5}=-6\sqrt{5}\)

1: =>|2x-1|=5

=>2x-1=5 hoặc 2x-1=-5

=>2x=6 hoặc 2x=-4

=>x=3 hoặc x=-2

2: \(\Leftrightarrow2\sqrt{x-3}+\dfrac{1}{3}\cdot3\sqrt{x-3}-\sqrt{x-3}=4\)

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

=>x-3=4

hay x=7

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

=>x-2=0 hoặc x+2=1

=>x=2 hoặc x=-1

1: \(=\sqrt{36}=6\)

2: \(=\sqrt{\left(15-9\right)\left(15+9\right)}=\sqrt{24\cdot6}=12\)

3: \(=3\sqrt{5}-1-3\sqrt{5}-1=-2\)

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

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

\(A=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\) ( ĐKXĐ : x \(\ge\) 2 )

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

\(A^2=2x+2\sqrt{x^2-8x+16}\)

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

\(A^2=2x+2|x-4|\)

\(\Rightarrow A=\sqrt{2x+2|x-4|}\)

A=\(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)

ĐKXĐ :x\(\ge2\)

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

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

* Nếu \(\sqrt{2}-\sqrt{x-2}\ge0\Leftrightarrow x\le4\) thì:

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

* Nếu \(\sqrt{2}-\sqrt{x-2}< 0\Leftrightarrow x>4\) thì :

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

\(A=\left(2-\sqrt{3}\right)\sqrt{4+2.2.\sqrt{3}+3}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=1\)

các câu còn lại làm tương tự nhé bạn !

a) 4x-3√x-1=4x-4√x+√x-1=4√x(√x-1)+(√x+1)

=(4√x+1).(√x-1)

c) 12-2√27=9-2.3√3+3=3²-2.3√3+(√3)²=(3-√3)²

d) 6x-√x-1=6x-3√x+2√x-1=3√x(2√x-1)+(2√x-1)

=(3√x+1).(2√x-1)

e) 11+2√30=6+2.√6.√5+5=(√6)²+2.√6.√5+(√5)²=(√6+√5)²

f) 10+2√21=7+2.√7.√3+3=(√7)²+2.√7.√3+(√3)²=(√7+√3)²