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

\(\sqrt{\dfrac{1}{9}}.0,04.64=\dfrac{1}{3}.\dfrac{64}{25}=\dfrac{64}{75}\)

Lời giải:

\(\sqrt{1991}=\sqrt{x}+\sqrt{y}\geq \sqrt{x}\Rightarrow x\leq 1991\)

Ta có:

\(\sqrt{x}+\sqrt{y}=\sqrt{1991}\Rightarrow \sqrt{y}=\sqrt{1991}-\sqrt{x}\)

\(\Rightarrow y=(\sqrt{1991}-\sqrt{x})^2=1991+x-2\sqrt{1991x}\)

\(\Rightarrow 2\sqrt{1991x}=1991+x-y\) là một số nguyên không âm.

Đặt \(2\sqrt{1991x}=t\) với $t$ là một số tự nhiên chẵn.

\(\Rightarrow 2^2(1991x)=t^2\)

\(\Rightarrow 1991x\) là số chính phương. Mà \(1991x=11.181.x\) nên $x$ phải có dạng \(11.181m^2\)

\(x\leq 1991\Rightarrow 11.181m^2\leq 1991\Rightarrow m^2\leq 1\Rightarrow m=\left\{0;1\right\}\)

+) $m=1$ thì $x=1991,y=0$

+) $m=0$ thì $x=0, y=1991$

^{a) Ta có:} \(\sqrt[3]{x^3+9x^2}=x+3< =>x^3+9x^2=x^3+9x^2+27x+27\)

Khi đó: \(27x+27=0=>x=-1\)

b) \(\sqrt[3]{5+x}=a;\sqrt[3]{5-x}=b\)

Khi đó ta có $HPT$:

\(a+b=1\) và \(a^3+b^3=10\)

Đến đây ez rồi nhé. Bạn chỉ cần giải hệ tìm $a;b$ sau đó thay vào tìm $x$ nhé!

\(\sqrt{17+12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)

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

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

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

\(\sqrt{17+12\sqrt{2}}+\sqrt{9+4\sqrt{2}}=\sqrt{\left(3+2\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}\)

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

\(\sqrt{17+12\sqrt{2}}+\sqrt{9+4\sqrt{2}}=\sqrt{8+2.2\sqrt{2}.3+9}+\sqrt{8+2.2\sqrt{2}.1+1}=\sqrt{\left(2\sqrt{2}+3\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}=\left|2\sqrt{2}+3\right|+\left|2\sqrt{2}+1\right|=2\sqrt{2}+3+2\sqrt{2}+1=4\sqrt{2}+4\)

\(a.\sqrt{0,4}.\sqrt{64}=\sqrt{0,4.64}=\sqrt{25,6}\)

\(b.\sqrt{5,2}.\sqrt{1,3}=\sqrt{5,2.1,3}=\sqrt{6,76}=2.6\)

\(c.\sqrt{12,1}.\sqrt{360}=\sqrt{12,1.360}=\sqrt{4356}=66\)

√(64.4)/100=16/10=1.6

√(13.13.4)=26

√(9.3.5.5)/10=15√3/√10=15√30/10

a) \(\sqrt{0,4}.\sqrt{64}=\dfrac{\sqrt{10}}{5}.8=\dfrac{8\sqrt{10}}{5}\)

b) \(\sqrt{5,2}.\sqrt{1,3}=\sqrt{5,2.1,3}=\sqrt{\dfrac{169}{25}}=\dfrac{13}{5}=2,6\)

c) \(\sqrt{12,1}.\sqrt{360}=\sqrt{12,1.360}=\sqrt{121.36}=\sqrt{121}.\sqrt{36}=11.6=66\)

B1: a) \(\sqrt{12,1.490}=\sqrt{12,1.10.49}=\sqrt{121}.\sqrt{49}=11.7=77\)

b) \(\sqrt{72.32}=\sqrt{36.2.32}=\sqrt{36}.\sqrt{64}=6.8=48\)

B2: a) \(\sqrt{48.75a^2}=\sqrt{3600a^2}=60\left|a\right|\)

b) \(\sqrt{8a^2}.\sqrt{18a^4}=\sqrt{8a^2.18a^4}=\sqrt{144a^6}=-12a^3\)

c) \(\sqrt{a}.\sqrt{\dfrac{9}{a}}=\sqrt{a.\dfrac{9}{a}}=\sqrt{9}=3\)

d) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)

a) \(\sqrt{\dfrac{1}{9}.0,4.64}=\sqrt{\dfrac{1}{9}}.\sqrt{0,4}.\sqrt{64}=\dfrac{1}{3}.\dfrac{\sqrt{10}}{5}.8=\dfrac{8\sqrt{10}}{15}\)

b) \(\sqrt{11\dfrac{1}{9}}=\sqrt{\dfrac{100}{9}}=\dfrac{\sqrt{100}}{\sqrt{9}}=\dfrac{10}{3}\)

c) \(\sqrt{\dfrac{1}{44}.2\dfrac{2}{49}}=\sqrt{\dfrac{1}{44}}.\sqrt{\dfrac{100}{49}}=\dfrac{\sqrt{11}}{22}.\dfrac{10}{7}=\dfrac{5\sqrt{11}}{77}\)

d) \(\sqrt{1\dfrac{9}{16}.2\dfrac{1}{4}.2\dfrac{7}{9}}\sqrt{\dfrac{25}{16}.\dfrac{9}{4}.\dfrac{25}{9}}=\sqrt{\dfrac{25}{16}}.\sqrt{\dfrac{9}{4}}.\sqrt{\dfrac{25}{9}}=\dfrac{5}{4}.\dfrac{3}{2}.\dfrac{5}{3}=\dfrac{25}{8}\)