Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 4:
a: a=1; b=-5; c=-7
Vì ac<0 nên phương trình có hai nghiệm trái dấu
b: \(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)
\(=5^2-2\cdot\left(-7\right)=25+14=39\)
\(\dfrac{1}{x_1^2}+\dfrac{1}{x_2^2}=\dfrac{x_1^2+x_2^2}{\left(x_1\cdot x_2\right)^2}=\dfrac{39}{7^2}=\dfrac{39}{49}\)
Bài 2: Để hpt có nghiệm duy nhất thì \(\dfrac{m}{1}\ne\dfrac{3}{-2}\Leftrightarrow\)\(m\ne\dfrac{-3}{2}\)
Bài 1: \(\left\{{}\begin{matrix}mx+y=5\left(1\right)\\2x-y=-2\left(2\right)\end{matrix}\right.\)
Lấy (1) cộng (2), ta được: \(\left(m+2\right)x=3\Rightarrow x=\dfrac{3}{m+2}\)
Thay vào (2): \(\dfrac{6}{m+2}-y=-2\)\(\Rightarrow y=\dfrac{6+2m+4}{m+2}=\dfrac{2m+10}{m+2}\)
x0+y0=1\(\Rightarrow\dfrac{3}{m+2}+\dfrac{2m+10}{m+2}=\dfrac{2m+13}{m+2}=1\)(ĐK: \(m\ne-2\))
\(\Rightarrow2m+13=m+2\Leftrightarrow m=-11\left(TM\right)\)
Bài 3: Thay \(x=\sqrt{2};y=4-\sqrt{2}\) vào đths y=ax+b:
\(\sqrt{2}a+b=4-\sqrt{2}\left(1\right)\)
Thay x=2; \(y=\sqrt{2}\) vào đths y=ax+b:
\(2a+b=\sqrt{2}\left(2\right)\)
Từ (1) và (2), ta có hpt: \(\left\{{}\begin{matrix}\sqrt{2}a+b=4-\sqrt{2}\\2a+b=\sqrt{2}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=-2\\b=\sqrt{2}+4\end{matrix}\right.\)
Vậy đths \(y=-2x+4+\sqrt{2}\) đi qua điểm \(\left(\sqrt{2};4-\sqrt{2}\right)\) và \(\left(2;\sqrt{2}\right).\)
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.
\(a.\sqrt{2a}.\sqrt{18a}=\sqrt{2a}.3\sqrt{2a}=3.2a=6a\)
\(b.\sqrt{3a.27ab^2}=\sqrt{9a^2b^2.9}=9\text{ |}ab\text{ |}\)
\(c.2y^2.\sqrt{\dfrac{x^4}{4y^2}}=2y^2.\dfrac{x^2}{-2y}=-x^2y\)
\(d.\dfrac{y}{x}.\sqrt{\dfrac{x^2}{y^4}}=\dfrac{y}{x}.\dfrac{x}{y^2}=\dfrac{1}{y}\)
\(e.\sqrt{\dfrac{9a^2}{16}}=\dfrac{3\text{ |}a\text{ |}}{4}\)
\(f.\sqrt{10.16a^2}=-4a\sqrt{10}\)
\(g.\sqrt{a^4\left(3-a\right)^2}=a^2\left(a-3\right)\)
\(h.\sqrt{\dfrac{2a^2b^4}{98}}\sqrt{\dfrac{a^2b^4}{49}}=\dfrac{b^2\text{ |}a\text{ |}}{7}\)
Câu 1:
\(\left\{{}\begin{matrix}x-2y=3-m\\4x+2y=6m+12\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=m+3\\y=m\end{matrix}\right.\)
\(\frac{x^2}{y+4}=\frac{5y+21}{x+1}\Leftrightarrow\frac{\left(m+3\right)^2}{m+4}=\frac{5m+21}{m+4}\) (\(m\ne-4\))
\(\Leftrightarrow m^2+6m+9=5m+21\)
\(\Leftrightarrow m^2+m-12=0\Rightarrow\left[{}\begin{matrix}m=3\\m=-4\end{matrix}\right.\)
Câu 2:
\(\Delta=m^2-8>0\Rightarrow m^2>8\)
Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=-m\\x_1x_2=2\end{matrix}\right.\)
\(x_1^2+x_2^2=\sqrt{5}^2\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=5\)
\(\Leftrightarrow m^2-4=5\)
\(\Rightarrow m^2=9\)
\(\Rightarrow m=\pm3\)
1.
a) \(\sqrt{3-2\sqrt{2}}+\sqrt{6-4\sqrt{2}}+\sqrt{9-4\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}+\sqrt{4-2.2.\sqrt{2}+2}+\sqrt{8-2.2\sqrt{2}.1+1}=\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}+\sqrt{2^2-2.2.\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}\right)^2-2.2\sqrt{2}.1+1^2}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|+\left|2-\sqrt{2}\right|+\left|2\sqrt{2}-1\right|=\sqrt{2}-1+2-\sqrt{2}+2\sqrt{2}-1=2\sqrt{2}\)
b) \(\sqrt{\left(4+\sqrt{10}\right)^2}-\sqrt{\left(4-\sqrt{10}\right)^2}=\left|4+\sqrt{10}\right|-\left|4-\sqrt{10}\right|=4+\sqrt{10}-4+\sqrt{10}=2\sqrt{10}\)
c) \(\dfrac{1}{\sqrt{2013}-\sqrt{2014}}-\dfrac{1}{\sqrt{2014}-\sqrt{2015}}=\dfrac{\sqrt{2013}+\sqrt{2014}}{\left(\sqrt{2013}-\sqrt{2014}\right)\left(\sqrt{2013}+\sqrt{2014}\right)}-\dfrac{\sqrt{2014}+\sqrt{2015}}{\left(\sqrt{2014}-\sqrt{2015}\right)\left(\sqrt{2014}+\sqrt{2015}\right)}=\dfrac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\dfrac{\sqrt{2014}+\sqrt{2015}}{2014-2015}=-\left(\sqrt{2013}+\sqrt{2014}\right)+\sqrt{2014}+\sqrt{2015}=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}+\sqrt{2015}=\sqrt{2015}-\sqrt{2013}\)
2.
a) \(x^2-2\sqrt{5}x+5=0\Leftrightarrow x^2-2.x.\sqrt{5}+\left(\sqrt{5}\right)^2=0\Leftrightarrow\left(x-\sqrt{5}\right)^2=0\Leftrightarrow x-\sqrt{5}=0\Leftrightarrow x=\sqrt{5}\)Vậy S={\(\sqrt{5}\)}
b) ĐK:x\(\ge-3\)
\(\sqrt{x+3}=1\Leftrightarrow\left(\sqrt{x+3}\right)^2=1^2\Leftrightarrow x+3=1\Leftrightarrow x=-2\left(tm\right)\)
Vậy S={-2}
3.
a) \(A=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
b) Ta có \(A=x-\sqrt{x}+1=x-2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2\ge0\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\Leftrightarrow A\ge\dfrac{3}{4}\)
Dấu bằng xảy ra khi x=\(\dfrac{1}{4}\)
Vậy GTNN của A=\(\dfrac{3}{4}\)
Lời giải:
Để PT có 2 nghiệm phân biệt $x_1,x_2$ thì:
\(\Delta'=(m+2)^2-(m^2+m+3)>0\)
\(\Leftrightarrow 3m+1>0\Leftrightarrow m> \frac{-1}{3}\)
Áp dụng định lý Vi-et: \(\left\{\begin{matrix} x_1+x_2=2(m+2)\\ x_1x_2=m^2+m+3\end{matrix}\right.\)
\(x_1x_2=m^2+m+3=(m+\frac{1}{2})^2+\frac{11}{4}\neq 0, \forall m>\frac{-1}{3}\) nên $x_1,x_2\neq 0$ với mọi \(m> \frac{-1}{3}\).
Khi đó:
\(\frac{x_1}{x_2}+\frac{x_2}{x_1}=1\)
\(\Leftrightarrow \frac{x_1^2+x_2^2}{x_1x_2}=4\)
\(\Leftrightarrow \frac{(x_1+x_2)^2-2x_1x_2}{x_1x_2}=4\)
\(\Leftrightarrow \frac{(x_1+x_2)^2}{x_1x_2}=6\Rightarrow (x_1+x_2)^2=6x_1x_2\)
\(\Leftrightarrow 4(m+2)^2=6(m^2+m+3)\)
\(\Leftrightarrow 2m^2-10m+2=0\)
\(\Leftrightarrow m=\frac{5\pm \sqrt{21}}{2}\) (thỏa mãn)
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: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
Câu 3: C
Câu 4: A
Câu 5: C
Câu 6: m=3
Câu 7: B
Câu 8: D
Câu 9: D
Câu 10: C