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b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
Bài 2:
a: \(=\sqrt{\left(\dfrac{1}{5a}\right)^2}=\dfrac{1}{\left|5a\right|}=\dfrac{-1}{5a}\)
b: \(=\dfrac{1}{3}\cdot15\cdot\left|a\right|=5\left|a\right|\)
Mọi ngươi giúp em với ạ chứ em làm câu a Bài 1 và 2 ra kết quả dài quá :(
Bài 1:
a: \(P=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b: Để P<1 thì P-1<0
\(\Leftrightarrow\dfrac{\sqrt{a}-4-\sqrt{a}+2}{\sqrt{a}-2}< 0\)
=>căn a-2>0
=>a>4
Bài 1
a) √81a - √36a - √144a = 9√a - 6√a - 12√a = -9√a
b) √75 - √48 - √300 = 5√3 - 4√3 - 10√3 = -9√3
Bài 2
a) √2x-3 = 7
⇒ 2x-3 = 49 ⇔ 2x = 52 ⇔ x =26
c) √16x - √9x = 2
⇔ 4√x - 3√x = 2 ⇔ √x = 2 ⇔ x = 4
Bài 3
a) √(2-√5)2 = l 2-√5 l = √5-2
b) (a - 3)2 + (a - 9)
= a2 - 6a + 9 + a - 9 = a2 - 5a
c) A=\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}:\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
=\(\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
=\(\left(\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\right)\)
=\(\left(\dfrac{-3\sqrt{x}-3}{x-9}\right).\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)
=\(\left(\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\right).\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)
=\(\dfrac{-3\sqrt{x}+9}{x-9}\)
Bài 2:
a: =>25x=35^2=1225
=>x=49
b: \(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
=>x+5=4
=>x=-1
a: \(=3xy\cdot\dfrac{\sqrt{2}}{\sqrt{xy}}=3\sqrt{2}\sqrt{xy}\)
b: \(=x\cdot\dfrac{\sqrt{6}}{\sqrt{x}}+\dfrac{\sqrt{6}}{3}\sqrt{x}\)
\(=\sqrt{6}\sqrt{x}+\dfrac{\sqrt{6}}{3}\sqrt{x}=\dfrac{4\sqrt{6}}{3}\cdot\sqrt{x}\)
c: \(=\sqrt{xy}+x\cdot\dfrac{\sqrt{y}}{\sqrt{x}}-y\cdot\dfrac{\sqrt{x}}{\sqrt{y}}\)
\(=\sqrt{xy}+\sqrt{xy}-\sqrt{xy}=\sqrt{xy}\)
đkxđ a>=0 a khác 1
\(C=\left(\frac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(C=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\sqrt{a}+3}{a-1}\)
\(C=\frac{\left(a-1\right).\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}\)
b)
\(a=4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)
\(\sqrt{a}=\sqrt{3}-1\)
thay vào nha
c) \(C=\frac{\left(a-1\right).\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}\)
để c<0 thì \(\frac{\left(a-1\right).\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}< 0\)
mà \(\sqrt{a}\left(\sqrt{a}+3\right)>0\)
\(\left(a-1\right)\left(\sqrt{a}+1\right)< 0\)
mà \(\sqrt{a}+1>0\)
nên a-1<0
\(0\le a< 1\)
Bài 1)
ĐK: \(x\geq 0; x\neq -4\)
Ta có:
\(A=\frac{1}{\sqrt{x}+2}+\frac{1}{2+\sqrt{x}}-\frac{2\sqrt{x}}{x+4}\)
\(=\frac{2}{\sqrt{x}+2}-\frac{2\sqrt{x}}{x+4}=2\left(\frac{1}{\sqrt{x}+2}-\frac{\sqrt{x}}{x+4}\right)\)
\(=2.\frac{x+4-x-2\sqrt{x}}{(\sqrt{x}+2)(x+4)}=2.\frac{4-2\sqrt{x}}{(\sqrt{x}+2)(x+4)}=\frac{4(2-\sqrt{x})}{(\sqrt{x}+2)(x+4)}\)
\(B=(\sqrt{2}+\sqrt{3}).\sqrt{2}-\sqrt{6}+\frac{\sqrt{333}}{\sqrt{111}}\)
\(=2+\sqrt{6}-\sqrt{6}+\frac{\sqrt{3}.\sqrt{111}}{\sqrt{111}}=2+\sqrt{3}\)
Để \(A=B\Leftrightarrow \frac{4(2-\sqrt{x})}{(\sqrt{x}+2)(x+4)}=2+\sqrt{3}\)
PT rất xấu. Mình nghĩ bạn đã chép sai biểu thức A.
Bài 2 : Tọa độ điểm B ?
Bài 3:
Để pt có hai nghiệm thì \(\Delta'=(m-3)^2-(m^2-1)>0\)
\(\Leftrightarrow 10-6m>0\Leftrightarrow m< \frac{5}{3}\)
Áp dụng định lý Viete: \(\left\{\begin{matrix} x_1+x_2=2(m-3)\\ x_1x_2=m^2-1\end{matrix}\right.\)
Khi đó:
\(4=2x_1+x_2=x_1+(x_1+x_2)=x_1+2(m-3)\)
\(\Rightarrow x_1=10-2m\)
\(\Rightarrow x_2=2(m-3)-(10-2m)=4m-16\)
Suy ra: \(\Rightarrow x_1x_2=(10-2m)(4m-16)\)
\(\Leftrightarrow m^2-1=8(5-m)(m-4)\)
\(\Leftrightarrow m^2-1=8(-m^2+9m-20)\)
\(\Leftrightarrow 9m^2-72m+159=0\)
\(\Leftrightarrow (3m-12)^2+15=0\) (vô lý)
Vậy không tồn tại $m$ thỏa mãn điều kiện trên.
3:
a: \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{6}{9}}=\dfrac{\sqrt{6}}{3}\)
b: \(\dfrac{x}{y}\cdot\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}\cdot\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{xy}}{y}\)
2:
a: 2căn 7=căn 28
3căn 2=căn 18
mà 28>18
nên 2*căn 7>3*căn 2
b: 5=2+3
mà 3>căn 2
nên 2+3>2+căn 2
=>5>2+căn 2
1) a) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)
\(=\sqrt{49.2}-\sqrt{36.2}+0,5\sqrt{4.2}\)
\(=7\sqrt{2}-6\sqrt{2}+0,5.2\sqrt{2}\)
\(=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
b) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49}\)
\(=3\sqrt{a}-4\sqrt{a}+7=7-\sqrt{a}\)
2. a) \(2\sqrt{7}=\sqrt{4.7}=\sqrt{28}\)
\(3\sqrt{2}=\sqrt{9.2}=\sqrt{18}\)
Mà \(\sqrt{28}>\sqrt{18}\Rightarrow2\sqrt{7}>3\sqrt{2}\)
b) \(5=2+3=2+\sqrt{9}\)
Vì \(\sqrt{9}>\sqrt{2}\Rightarrow2+\sqrt{9}>2+\sqrt{2}\Rightarrow5>2+\sqrt{2}\)
3. a) \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{6}{9}}=\dfrac{\sqrt{6}}{3}\)
b) \(\dfrac{x}{y}.\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}.\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{xy}}{y}\)