Chứng minh hằng đẳng thức
a) \(\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}=\sqrt{2\left(a+\sqrt{a^2}-b\right)}\)
b)\(\sqrt{a+\sqrt{b}}-\sqrt{a-\sqrt{b}}=\sqrt{2\left(a-\sqrt{a^2-b}\right)}\)
Chứng minh đẳng thức sau với \(b\ge0;a\ge\sqrt{b}\)
\(\sqrt{a+\sqrt{b}}\mp\sqrt{a-\sqrt{b}}=\sqrt{2\left(a\mp\sqrt{a^2-b}\right)}\)
\(\Rightarrow\left(\sqrt{a+\sqrt{b}}\mp\sqrt{a-\sqrt{b}}\right)^2=\left(\sqrt{2\left(a\mp\sqrt{a^2-b}\right)}\right)^2\Leftrightarrow a+\sqrt{b}+a-\sqrt{b}\mp2\sqrt{\left(a+\sqrt{b}\right)\cdot\left(a-\sqrt{b}\right)}=2a\mp2\sqrt{a^2-b}\Leftrightarrow2a\mp2\sqrt{a^2-b}=2a\mp2\sqrt{a^2-b}\) (luôn đúng) \(\Rightarrowđpcm\)
1. chứng minh rằng các hằng đẳng thức sau với điều kiện các biểu thức tồn tại:
a) \(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}=a-b\)
b)\(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a\)
a, \(VT=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)=a-b=VP\) đpcm
b,\(VT=1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}-\dfrac{a^2-a}{a-1}=1-\sqrt{a}+\sqrt{a}-a=1-a=VP\) đpcm
Chứng minh các đẳng thức sau:
a) \(\left(1-a^2\right):\left(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\right)+1=\frac{2}{1-a}\)
b) \(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{ab}+b}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)=\sqrt{b}-\sqrt{a}\)
c) \(\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)=\frac{\sqrt{a}}{a}\)
Chứng minh các đẳng thức sau:
a) \(\left(1-a^2\right):\left[\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1
+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\right]+1=\frac{2}{1-a}\)
b) \(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{ab}+b}
+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)=\sqrt{b}-\sqrt{a}\)
c) \(\frac{\sqrt{a}+\sqrt{b}-1}{a
+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a
+\sqrt{ab}}\right)=\frac{\sqrt{a}}{a}\)
d) \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\)
Tính giá trị biểu thức (Nhân thêm số căn vào biểu thức để làm xuất hiện hằng đẳng thức \(\left(a\pm\sqrt{b}\right)^2\) hoặc \(\left(\sqrt{a}\pm\sqrt{b}\right)^2\) rồi phá căn)
a. \(\left(4\sqrt{2}+\sqrt{30}\right).\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)
b. \(\dfrac{\sqrt{3}+1}{2}.\sqrt{8-2\sqrt{3}}\)
a) \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right).\sqrt{\dfrac{8-2\sqrt{15}}{2}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{25.6}-\sqrt{9.10}\right).\sqrt{\dfrac{\left(\sqrt{5}\right)^2-2\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}{2}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right).\sqrt{\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{2}}\)
\(=\left(\sqrt{10}+\sqrt{6}\right).\dfrac{\left|\sqrt{5}-\sqrt{3}\right|}{\sqrt{2}}=\sqrt{2}.\left(\sqrt{5}+\sqrt{3}\right).\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=2\)
a) Ta có: \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\sqrt{8-2\sqrt{15}}\cdot\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{5}-\sqrt{3}\right)^2\cdot\left(4+\sqrt{15}\right)\)
\(=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\)
\(=32+8\sqrt{15}-8\sqrt{15}-30\)
=2
Chứng minh các hằng đẳng thức sau với \(b\ge9,a\ge\sqrt{b}\)
a, \(\sqrt{a+\sqrt{b}}\pm\sqrt{a-\sqrt{b}}=\sqrt{2\left(a\pm\sqrt{a^2-b}\right)}\)
b, \(\sqrt{a\pm\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\pm\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\)
Chứng minh đẳng thức:
\(\dfrac{a\sqrt{b}+b}{a-b}.\sqrt{\dfrac{ab+b^2-2\sqrt{ab^2}}{a\left(a+2\sqrt{b}+b\right)}}\left(\sqrt{a}+\sqrt{b}\right)=b\) (với a > b > 0)
Chứng minh các đẳng thức (với a, b không âm và \(a\ne b\))
a) \(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
b) \(\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}\right)\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\)
a) \(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}\)
=\(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
=\(\dfrac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
=\(\dfrac{4\sqrt{ab}+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\dfrac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
=\(\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)(đpcm)
a) Ta có:
\(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}-\dfrac{2b}{b-a}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
(với a, b không âm và a ≠b )
Vế trái bằng vế phải nên đẳng thức được chứng minh.
b) Ta có:
\(\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
\(=\left(\dfrac{\sqrt{a^3}+\sqrt{b^3}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left[\dfrac{\sqrt{a}+\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right]^2\)
\(=\left[\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a^2}-\sqrt{ab}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right]\left(\dfrac{1}{\sqrt{a}-\sqrt{b}}\right)^2\)
\(=\left(\sqrt{a^2}-\sqrt{ab}+\sqrt{b^2}-\sqrt{ab}\right)\dfrac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}=1\)
(với a, b không âm và a ≠b )
Vế trái bằng vế phải nên đẳng thức được chứng minh.
ba số dương a,b,c thỏa mãn \(b\ne c,\sqrt{a}+\sqrt{b}\ne\sqrt{c}\) và\(a+b=\left(\sqrt{a}+\sqrt{b}-\sqrt{c}\right)^2\).chứng minh đẳng thức
\(\dfrac{a+\left(\sqrt{a}-\sqrt{c}\right)^2}{b+\left(\sqrt{b}-\sqrt{c}\right)^2}=\dfrac{\sqrt{a}-\sqrt{c}}{\sqrt{b}-\sqrt{c}}\)
Lời giải:
Đặt \((\sqrt{a}, \sqrt{b}, \sqrt{c})=(x,y,z)\). Bài toán trở thành
Cho $x,y,z$ dương thỏa mãn \(y^2\neq z^2; x+y\neq z; x^2+y^2=(x+y-z)^2\)
CMR: \(\frac{x^2+(x-z)^2}{y^2+(y-z)^2}=\frac{x-z}{y-z}\)
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Ta có:
\(x^2+y^2=(x+y-z)^2=[y+(x-z)]^2\)
\(\Leftrightarrow x^2+y^2=y^2+(x-z)^2+2y(x-z)\)
\(\Leftrightarrow x^2=(x-z)^2+2y(x-z)\)
\(\Leftrightarrow x^2+(x-z)^2=2(x-z)^2+2y(x-z)=2(x-z)(x-z+y)\)
Tương tự:
\(y^2+(y-z)^2=2(y-z)^2+2x(y-z)=2(y-z)(y-z+x)\)
Do đó: \(\frac{x^2+(x-z)^2}{y^2+(y-z)^2}=\frac{2(x-z)(x-z+y)}{2(y-z)(y-z+x)}=\frac{x-z}{y-z}\)
Ta có đpcm.