Bài 75 (trang 40 SGK Toán 9 Tập 1)
Chứng minh các đẳng thức sau:
a) $\left(\dfrac{2 \sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}=-1,5$;
b) $\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right): \dfrac{1}{\sqrt{7}-\sqrt{5}}=-2$;
c) $\dfrac{a \sqrt{b}+b \sqrt{a}}{\sqrt{a b}}: \dfrac{1}{\sqrt{a}-\sqrt{b}}=a-b$ với $a, b$ dương và $a \neq b$;
d)...
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Bài 75 (trang 40 SGK Toán 9 Tập 1)
Chứng minh các đẳng thức sau:
a) $\left(\dfrac{2 \sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}=-1,5$;
b) $\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right): \dfrac{1}{\sqrt{7}-\sqrt{5}}=-2$;
c) $\dfrac{a \sqrt{b}+b \sqrt{a}}{\sqrt{a b}}: \dfrac{1}{\sqrt{a}-\sqrt{b}}=a-b$ với $a, b$ dương và $a \neq b$;
d) $\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a$ với $a \geq 0$ và $a \neq 1$.
a) VT=\left(\dfrac{2 \sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}VT=(8−223−6−3216)⋅61
=\left(\dfrac{\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{3}-\sqrt{6}}{\sqrt{2^{2} \cdot 2}-2}-\dfrac{\sqrt{6^{2} .6}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}=(22⋅2−22⋅2⋅3−6−362.6)⋅61
=\left(\dfrac{\sqrt{2} \cdot \sqrt{6}-\sqrt{6}}{2 \sqrt{2}-2}-\dfrac{6 . \sqrt{6}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}=(22−22⋅6−6−36.6)⋅61
=\left[\dfrac{\sqrt{6}(\sqrt{2}-1)}{2(\sqrt{2}-1)}-\dfrac{6 \sqrt{6}}{3}\right] \cdot \dfrac{1}{\sqrt{6}}=[2(2−1)6(2−1)−366]⋅61
=\left(\dfrac{\sqrt{6}}{2}-2 \sqrt{6}\right) \cdot \dfrac{1}{\sqrt{6}}=(26−26)⋅61
=\left(\dfrac{\sqrt{6}}{2}-\dfrac{4 \sqrt{6}}{2}\right) \cdot \dfrac{1}{\sqrt{6}}=(26−246)⋅61
=\left(\dfrac{-3}{2} \sqrt{6}\right) \cdot \dfrac{1}{\sqrt{6}}=(2−36)⋅61
=-\dfrac{3}{2}=-1,5=V P=−23=−1,5=VP.
b) VT=\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right): \dfrac{1}{\sqrt{7}-\sqrt{5}}VT=(1−214−7+1−315−5):7−51
=\left(\dfrac{\sqrt{7} \cdot \sqrt{2}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{5} \cdot \sqrt{3}-\sqrt{5}}{1-\sqrt{3}}\right): \dfrac{1}{\sqrt{7}-\sqrt{5}}=(1−27⋅2−7+1−35⋅3−5):7−51
=\left[\dfrac{\sqrt{7}(\sqrt{2}-1)}{1-\sqrt{2}}+\dfrac{\sqrt{5}(\sqrt{3}-1)}{1-\sqrt{3}}\right]: \dfrac{1}{\sqrt{7}-\sqrt{5}}=[1−27(2−1)+1−35(3−1)]:7−51
=(-\sqrt{7}-\sqrt{5})(\sqrt{7}-\sqrt{5})=(−7−5)(7−5)
=-(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})=−(7+5)(7−5)
=-(7-5)=-2=VP=−(7−5)=−2=VP.
c) V T=\dfrac{a \sqrt{b}+b \sqrt{a}}{\sqrt{a b}}: \dfrac{1}{\sqrt{a}-\sqrt{b}}VT=abab+ba:a−b1
=\dfrac{\sqrt{a} \cdot \sqrt{a} \cdot \sqrt{b}+\sqrt{b} \cdot \sqrt{b} \cdot \sqrt{a}}{\sqrt{a b}}: \dfrac{1}{\sqrt{a}-\sqrt{b}}=aba⋅a⋅b+b⋅b⋅a:a−b1
=\dfrac{\sqrt{a} \cdot \sqrt{a b}+\sqrt{b} \cdot \sqrt{a b}}{\sqrt{a b}}: \dfrac{1}{\sqrt{a}-\sqrt{b}}=aba⋅ab+b⋅ab:a−b1
=\dfrac{\sqrt{a b}(\sqrt{a}+\sqrt{b})}{\sqrt{a b}} \cdot(\sqrt{a}-\sqrt{b})=abab(a+b)⋅(a−b)
=(\sqrt{a}+\sqrt{b}) \cdot(\sqrt{a}-\sqrt{b})=(a+b)⋅(a−b)
=a-b=V P=a−b=VP.
d) VT=\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)VT=(1+a+1a+a)(1−a−1a−a)
=\left(1+\dfrac{\sqrt{a} \cdot \sqrt{a}+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{\sqrt{a} \cdot \sqrt{a}-\sqrt{a}}{\sqrt{a}-1}\right)=(1+a+1a⋅a+a)(1−a−1a⋅a−a)
=\left[1+\dfrac{\sqrt{a}(\sqrt{a}+1)}{\sqrt{a}+1}\right]\left[1-\dfrac{\sqrt{a}(\sqrt{a}-1)}{\sqrt{a}-1}\right]=[1+a+1a(a+1)][1−a−1a(a−1)]
=(1+\sqrt{a})(1-\sqrt{a})=(1+a)(1−a)
=1-(\sqrt{a})^{2}=1-a=V P=1−(a)2=1−a=VP