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f) \(\left(\sqrt{6x+1}-\sqrt{6x-1}\right)^2=\left(\sqrt{6x+1}\right)^2-2\sqrt{\left(6x+1\right)\left(6x-1\right)}+\left(\sqrt{6x-1}\right)^2\)
\(=6x+1+6x-1-2\sqrt{36x^2-1}=12x-2\sqrt{36x^2-1}\)
tương tự các câu khác mình làm tắt chút nha:
c) \(\left(\sqrt{2x+3}+\sqrt{2x-3}\right)^2=2x+3+2x-3-2\sqrt{\left(2x+3\right)\left(2x-3\right)}=4x+2\sqrt{4x^2-9}\)
d) \(\left(\sqrt{2x+y}+\sqrt{2x-y}\right)^2=2x+y+2x-y-2\sqrt{\left(2x+y\right)\left(2x-y\right)}=4x-2\sqrt{4x^2-y^2}\)
\(\left(\sqrt{5x-2}-\sqrt{5x+2}\right)^2=5x-2+5x+2-2\sqrt{\left(5x-2\right)\left(5x+2\right)}=10x-2\sqrt{25x^2-4}\)
A\(=\left|5x-1\right|-\left|6x\right|\)
TH1: x<0
A=1-5x+6x=x+1
TH2: 0<=x<1/5
=>A=1-5x-6x=1-11x
TH3: x>=1/5
A=5x-1-6x=-x-1
\(A=\left(\dfrac{6x+4}{3\sqrt{3x^3}-8}-\dfrac{\sqrt{3x}}{3x+2\sqrt{3x}+4}\right).\left(\dfrac{1+3\sqrt{3x^3}}{1+\sqrt{3x}}-\sqrt{3x}\right)\)
Điều kiện tự làm nha:
Đặt \(\sqrt{3x}=a\) thì ta có:
\(A=\left(\dfrac{2a^2+4}{a^3-8}-\dfrac{a}{a^2+2a+4}\right).\left(\dfrac{1+a^3}{1+a}-a\right)\)
\(=\left(\dfrac{2a^2+4}{\left(a-2\right)\left(a^2+2a+4\right)}-\dfrac{a}{a^2+2a+4}\right).\left(\dfrac{\left(1+a\right)\left(1-a+a^2\right)}{1+a}-a\right)\)
\(=\dfrac{a^2+2a+4}{\left(a-2\right)\left(a^2+2a+4\right)}.\left(1-2a+a^2\right)\)
\(=\dfrac{\left(a-1\right)^2}{a-2}=\dfrac{\left(\sqrt{3x}-1\right)^2}{\sqrt{3x}-2}\)
A)\(\left(\sqrt{5-2}+\sqrt{5+2}\right)^2=\left(\sqrt{5-2}\right)^2+2\sqrt{5-2}\sqrt{5+2}+\left(\sqrt{5-2}\right)^2\)\(=5-2+6+5+2=16\)
B)\(\left(\sqrt{x+y}-\sqrt{x-y}\right)^2=\left(\sqrt{x+y}\right)^2-2\sqrt{x-y}\sqrt{x+y}+\left(\sqrt{x-y}\right)2\)
\(=x+y-2x+2y+x-y=2y\), Cho mik đúng nha bn!
\(\left(\sqrt{6x+1}-\sqrt{6x-1}\right)^2=\left(\sqrt{6x+1}\right)^2-2\sqrt{\left(6x+1\right)\left(6x-1\right)}+\left(\sqrt{6x-1}\right)^2\)
\(=6x+1+6x-1-2\sqrt{36x^2-1}=12x-2\sqrt{36x^2-1}\)
\(\left(\sqrt{5x-2}-\sqrt{5x+2}\right)^2=5x-2+5x+2-2\sqrt{\left(5x-2\right)\left(5x+2\right)}=10x-2\sqrt{25x^2-4}\)