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

1, \(\sqrt{8}-3\sqrt{32}+\sqrt{72}=2\sqrt{2}-12\sqrt{2}+6\sqrt{2}=-4\sqrt{2}\)

2,\(6\sqrt{12}-2\sqrt{48}+5\sqrt{75}-7\sqrt{108}=12\sqrt{3}-8\sqrt{3}+25\sqrt{3}-42\sqrt{3}=-13\sqrt{3}\)

3, \(\sqrt{20}+3\sqrt{45}-6\sqrt{80}-\dfrac{1}{3}\sqrt{125}=2\sqrt{5}+9\sqrt{5}-24\sqrt{5}-\dfrac{5}{3}.\sqrt{5}=-\dfrac{44}{3}.\sqrt{5}\)

4, \(2\sqrt{5}-\sqrt{125}-\sqrt{80}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}=-7\sqrt{5}\)

5, \(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}=3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}=-10\sqrt{2}\)

Câu a : \(\sqrt{40^2-24^2}=\sqrt{\left(40+24\right)\left(40-24\right)}=\sqrt{64.16}=8.4=32\)

Câu b : \(\sqrt{52^2-48^2}=\sqrt{\left(52+48\right)\left(52-48\right)}=\sqrt{100.4}=10.2=20\)

\(\left(2\sqrt{2}-\sqrt{3}\right)^2=8-2\cdot2\sqrt{2}\cdot\sqrt{3}+3=11-4\sqrt{6}\)

\(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2=3-\sqrt{5}+3+\sqrt{5}+2\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}=6+2\sqrt{9-5}=6+2.2=10\)

a.\(\Leftrightarrow\left(\sqrt{4+\sqrt{15}}\right)^2\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)

\(\Leftrightarrow\sqrt{2}\sqrt{4+\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}\)

\(\Leftrightarrow\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4^2-15}\)

\(\Leftrightarrow\sqrt{5+2\sqrt{5.3}+3}\left(\sqrt{5}-\sqrt{3}\right)\)

\(\Leftrightarrow\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\)

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

b.T\(^2\)câu a ta tách (3+\(\sqrt{5}\))=\(\left(\sqrt{3+\sqrt{5}}\right)^2\);(\(\sqrt{10}-\sqrt{6}\))\(=\sqrt{2}\left(\sqrt{5}-1\right)\)

Rồi tính tiếp với đáp số =8

c.\(\Leftrightarrow\dfrac{\sqrt{2}\sqrt{\sqrt{5}+1}\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)}{\sqrt{2}\sqrt{\sqrt{5}+1}\sqrt{\sqrt{5}+1}}-\sqrt{2-2\sqrt{2}+1}\)

\(\Leftrightarrow\dfrac{\sqrt{2\left(\sqrt{5}+1\right)\left(\sqrt{5}+2\right)}+\sqrt{\left(2\sqrt{5}+1\right)\left(\sqrt{5}-2\right)}}{\sqrt{2}(\sqrt{5}+1)}-\sqrt{\left(\sqrt{2}-1\right)^2}\)

\(\Leftrightarrow\dfrac{\sqrt{14+6\sqrt{5}}+\sqrt{6-2\sqrt{5}}}{\sqrt{2}\left(\sqrt{5}+1\right)}-\left(\sqrt{2}-1\right)\)

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

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

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

27M² = 27(a²b²c²) \(\le\)(a² + b² + c²)³ = 225³

\(\Rightarrow\)M² \(\le\)75³

\(\Leftrightarrow-375\sqrt{3}\le M\le375\sqrt{3}\)

\(\sqrt{2x+5}=5\left(x\ge-\dfrac{5}{2}\right)\)

\(\Leftrightarrow2x+5=25\)

\(\Leftrightarrow2x=20\)

\(\Leftrightarrow x=10\left(n\right)\)

\(\sqrt{2x+5}=5\left(x\ge-\dfrac{5}{2}\right)\)

\(\Leftrightarrow2x+5=25\Leftrightarrow2x=20\Leftrightarrow x=10\left(TM\right)\)

KL.......

\(\sqrt{2x+5}=5\)

Như này à bn?

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

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

\(=\dfrac{3\sqrt{3}+6}{3}+\dfrac{\left(2+\sqrt{2}\right)\sqrt{3}}{3}-\sqrt{2}-\sqrt{3}\)

\(=\dfrac{3+\left(\sqrt{3}+2\right)}{3}+\dfrac{2\sqrt{3}+\sqrt{6}}{3}-\sqrt{2}-\sqrt{3}\)

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

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

\(A=\dfrac{u-v}{\sqrt{u}+\sqrt{v}}-\dfrac{\sqrt{u^3}+\sqrt{v^3}}{u-v}\)

\(=\sqrt{u}-\sqrt{v}-\dfrac{u\sqrt{u}+v\sqrt{v}}{\left(\sqrt{u}-\sqrt{v}\right)\left(\sqrt{u}+\sqrt{v}\right)}\)

\(=\sqrt{u}-\sqrt{v}-\dfrac{u-\sqrt{uv}+v}{\left(\sqrt{u}-\sqrt{v}\right)\left(\sqrt{u}+\sqrt{v}\right)}\)

\(=\sqrt{u}-\sqrt{v}-\dfrac{u-\sqrt{uv}+v}{\sqrt{u}-\sqrt{v}}\)

\(=\dfrac{\left(\sqrt{u}-\sqrt{v}\right)\sqrt{u}-\left(\sqrt{u}-\sqrt[]{v}\right)\sqrt{v}-\left(u-\sqrt{uv}+v\right)}{\sqrt{u}-\sqrt{v}}\)

\(=\dfrac{u-\sqrt{uv}-\sqrt{uv}+v-u+\sqrt{uv}-v}{\sqrt{u}-\sqrt{v}}\)

\(\Leftrightarrow\)\(-\dfrac{\sqrt{uv}}{\sqrt{u}-\sqrt{v}}\)

để cả căn hơi phức tạp nhỉ? nếu tinh ý 1 chút thì sẽ đơn giản thôi :3

chú ý nhé ! nếu ta đăt như sau \(\sqrt{u}=a;\sqrt{v}=b\)

đến đấy thì dễ nhỉ<3;

\(A=\dfrac{a^2-b^2}{a+b}-\dfrac{a^3+b^3}{a^2-b^2}\)

xem nào ~~ để ý xem nó có phải hằng đẳng thức quen thuộc k nhỉ, thôi k quan tâm cứ trâu bò vào xem ra cái j k đã bạn ạ

\(A=\dfrac{\left(a+b\right)\left(a-b\right)}{a+b}-\dfrac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a-b\right)\left(a+b\right)}\)

\(A=a-b-\dfrac{a^2-ab+b^2}{a-b}\) có thể bạn nghĩ đến đây là khó, đùng ngại ta hãy cứ quy đồng chúng

\(A=\dfrac{\left(a-b\right)^2-a^2+ab-b^2}{a-b}=\dfrac{-ab}{a-b}\)