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\(1a.A=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}=\dfrac{6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{3}=\dfrac{2}{\sqrt{x}+3}\) ( x ≥ 0 ; x # 9 )
\(b.A>\dfrac{1}{3}\) ⇔ \(\dfrac{2}{\sqrt{x}+3}>\dfrac{1}{3}\text{⇔}\dfrac{3-\sqrt{x}}{3\left(\sqrt{x}+3\right)}>0\)
⇔ \(3-\sqrt{x}>0\)
⇔ \(x< 9\)
Kết hợp ĐKXĐ , ta có : \(0\text{≤}x< 9\)
\(c.\) Tìm GTLN chứ ?
\(A=\dfrac{2}{\sqrt{x}+3}\text{≤}\dfrac{2}{3}\)
⇒ \(A_{MAX}=\dfrac{2}{3}."="x=0\left(TM\right)\)
\(a.VT=2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}=9=VP\)Vậy , đẳng thức được chứng minh .
\(b.VT=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3+2\sqrt{3}+1}+\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}=VP\)Vậy , đẳng thức được chứng minh .
\(c.VT=\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}=\dfrac{2\left(\sqrt{5}+2\right)-2\left(\sqrt{5}-2\right)}{5-4}=8=VP\)Vậy , đẳng thức được chứng minh .
=\(\sqrt{3-\sqrt{5}}\)\(\sqrt{2}\)(\(\sqrt{5}-1\)) (\(3+\sqrt{5}\))
=\(\sqrt{6-2\sqrt{5}}\)(\(\sqrt{5}-1\)) (\(3+\sqrt{5}\))
=\(\sqrt{\left(\sqrt{5}+1\right)^2}\)(\(\sqrt{5}-1\))(\(3+\sqrt{5}\))
=(\(\sqrt{5}+1\))(\(\sqrt{5}-1\))(\(3+\sqrt{5}\))
=4(\(3+\sqrt{5}\))
=12+4\(\sqrt{5}\)
1. \(\left(\sqrt{5}-\sqrt{6}\right)=\left(\sqrt{5}\right)^2-2\sqrt{5}\sqrt{6}+\left(\sqrt{6}\right)^2=5-2\sqrt{30}+6\)
2. \(\left(\sqrt{3}-\sqrt{5}\right)^2=\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot\sqrt{5}+\left(\sqrt{5}\right)^2=3-2\sqrt{15}+5\)
3. \(\left(2\sqrt{2}+\sqrt{3}\right)^2=\left(2\sqrt{2}\right)^2+2\cdot2\sqrt{2}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2=8+4\sqrt{6}+3\)
4. \(\left(\sqrt{4}-\sqrt{17}\right)^2=\left(\sqrt{4}\right)^2-2\cdot\sqrt{4}\cdot\sqrt{17}+\left(\sqrt{17}\right)^2=4-4\sqrt{47}+17\)
5. \(\sqrt{\left(\sqrt{5}-3\right)^2}=\left|\sqrt{5}-3\right|=\left|-3+\sqrt{5}\right|=3-\sqrt{5}\)
6. \(\left(2\sqrt{5}-\sqrt{7}\right)\left(2\sqrt{5}+\sqrt{7}\right)=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=4\cdot5-7=13\)
7. \(\left(5\sqrt{2}+2\sqrt{3}\right)\left(2\sqrt{3}-5\sqrt{2}\right)=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)
8. \(\sqrt{\left(5+2\sqrt{6}\right)^2}-\sqrt{\left(5-2\sqrt{6}\right)^2}=\left|5+2\sqrt{6}\right|-\left|5-2\sqrt{6}\right|=5+2\sqrt{6}-\left(5-2\sqrt{6}\right)=4\sqrt{6}\)9. \(\sqrt{\left(\sqrt{7}-2\right)^2}+\sqrt{\left(\sqrt{7}+2\right)^2}=\left|\sqrt{7}-2\right|+\left|\sqrt{7}+2\right|=-2+\sqrt{7}+2+\sqrt{7}=2\sqrt{7}\)
10. \(\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\left|\sqrt{3}+\sqrt{2}\right|+\left|\sqrt{3}-\sqrt{2}\right|=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)
#em mới lớp 8 nên không chắc lắm ạ :((
L
nhờ vào năng lực rinegan , ta có thể đoán dc
\(\left(\sqrt{1+x}+\sqrt{8-x}\right)^2=1+x+8-x-2\sqrt{\left(X+1\right)\left(8-x\right)}\)
vậy pt sẽ như sau
\(a,\left(\sqrt{1+x}+\sqrt{8-x}\right)^2-\sqrt{\left(1+x\right)\left(8-x\right)}=3\) " thêm bớt nếu m thông minh sẽ hiểu "
\(9+2\sqrt{\left(1+x\right)\left(8-x\right)}-\sqrt{\left(1+x\right)\left(8-x\right)}=3\)
\(\sqrt{\left(1+x\right)\left(8-x\right)}=-6\)
\(\left(1+x\right)\left(8-x\right)=36\)
đến đây m có thể tự làm
c) \(\sqrt{x+5}=5-x^2\)
\(x+5=\left(5-x\right)^2\)
\(x+5=x^4-10x^2+25\) " rồi xong pt bậc 4 :)
\(x^4-10x^2-x+20=0\)
\(x^4=10x^2+x-20\)
\(x^4+2mx^2+m^2=10x^2+x-20+2mx^2+m^2\)
\(\left(x^2+m\right)^2=2x^2\left(5+m\right)+x+\left(m^2-20\right)\)
\(\Delta=1-8\left(5+m\right)\left(m^2-20\right)\)
\(\Delta=1-8\left(5m^2-100+m^3-20m\right)\)
\(\Delta=1-40m^2+800-8m^3+160m\)
\(\Delta=-\left(2m+9\right)\left(4m^2+2m-89\right)\)
lấy m= -9/2 , cho nhanh thay vào ta đươc
\(\left(x^2-\frac{9}{2}\right)^2=2x^2\left(5-\frac{9}{2}\right)+x+\left(\frac{9}{2}^2-20\right)\)
\(\left(x^2-\frac{9}{2}\right)^2=x^2+x+\frac{1}{4}\)
\(\left(x^2-\frac{9}{2}\right)^2=\left(x+\frac{1}{2}\right)^2\)
\(\hept{\begin{cases}x^2-\frac{9}{2}=x+\frac{1}{2}\\x^2-\frac{9}{2}=-x-\frac{1}{2}\end{cases}}\)
đến đây cậu có thể làm tiếp :)
câu B hơi gắt cần time suy nghĩ :)
\(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right).\sqrt{3-\sqrt{5}}\)
= \(\left(3+\sqrt{5}\right).\left(\sqrt{5}-1\right).\sqrt{2}.\sqrt{3-\sqrt{5}}\)
= \(\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}\)
= \(\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right).\sqrt{\left(\sqrt{5-1}\right)^2}\)
= \(\left(3+\sqrt{5}\right).\left(\sqrt{5}-1\right)^2\)
= \(\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)\)
= \(2.\left(3+\sqrt{5}\right).\left(3-\sqrt{5}\right)\)
= \(2.\left(9-5\right)\)
= \(2.4=8\)
Chúc bạn học tốt !!!
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