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\(P=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\) (Đk:\(a>0\))
\(=\dfrac{\sqrt{a}\left(a\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=\sqrt{a}\left(\sqrt{a}+1\right)-2\sqrt{a}-1+1\)
\(=a-\sqrt{a}\)
b) \(P=2\Leftrightarrow a-\sqrt{a}=2\Leftrightarrow a-\sqrt{a}-2=0\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}=2\\\sqrt{a}=-1\left(vn\right)\end{matrix}\right.\)\(\Rightarrow a=4\) (tm)
Vậy a=4 thì P=2
c) \(P=a-\sqrt{a}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\)
Vậy \(P_{min}=-\dfrac{1}{4}\)
Coi pt \(a-\sqrt{a}-2=0\) là pt ẩn \(\sqrt{a}\)
Hoặc e đặt \(t=\sqrt{a}\)
Pt tt: \(t^2-t-2=0\) \(\Leftrightarrow\left[{}\begin{matrix}t=-1\\t=2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}=-1\\\sqrt{a}=2\end{matrix}\right.\)
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2:
1+cot^2a=1/sin^2a
=>1/sin^2a=1681/81
=>sin^2a=81/1681
=>sin a=9/41
=>cosa=40/41
tan a=1:40/9=9/40
![](https://rs.olm.vn/images/avt/0.png?1311)
sin 650=cos 350
\(cos70^0=sin30^0\)
\(tan80^0=cot20^0\)
\(cot68^0=tan32^0\)
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a: \(\Leftrightarrow\left\{{}\begin{matrix}3x-2y=-1\\8x-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5x=-5\\4x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}4x+8y=-4\\4x+3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5y=-5\\x+2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}3x-6y=-12\\-3x+6y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0x=-2\left(loại\right)\\-3x+6y=10\end{matrix}\right.\Leftrightarrow\left(x,y\right)\in\varnothing\)
d: \(\Leftrightarrow\left\{{}\begin{matrix}2x+y=-2\\2x+y=-2\end{matrix}\right.\Leftrightarrow\left(x,y\right)\in R\)
\(a,\left\{{}\begin{matrix}3x-2y=-1\\4x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2\left(4x-2\right)=-1\\y=4x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-8x+4=-1\\y=4x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5x=-5\\y=4x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=4.1-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
\(b,\left\{{}\begin{matrix}x+2y=-1\\4x+3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1-2y\\4\left(-1-2y\right)+3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1-2y\\-4-8y+3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1-2y\\-5y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1-2\left(-1\right)\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,\left\{{}\begin{matrix}x-2y=-4\\-3x+6y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-4\\-3\left(2y-4\right)+6y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-4\\-6y+12+6y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-4\\12=10\left(vô.lí\right)\end{matrix}\right.\)
\(d,\left\{{}\begin{matrix}2x+y=-2\\4x+2y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y=-2\\2x+y=-2\left(luôn.đúng\right)\end{matrix}\right.\)
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Bài 3:
a) Ta có: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{3}{\sqrt{x}-3}\right)\cdot\dfrac{\sqrt{x}+3}{x+9}\)
\(=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\)
\(=\dfrac{1}{\sqrt{x}-3}\)
b) Ta có: \(B=21\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)^2-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)^2-15\sqrt{15}\)
\(=21\left(5+\sqrt{3}-\sqrt{5}+2\sqrt{\left(2+\sqrt{3}\right)\left(3-\sqrt{5}\right)}\right)-6\left(5-\sqrt{3}+\sqrt{5}+2\sqrt{\left(2-\sqrt{3}\right)\left(3+\sqrt{5}\right)}\right)-15\sqrt{15}\)
\(=21\left(4+\sqrt{15}\right)-6\left(4+\sqrt{15}\right)-15\sqrt{15}\)
\(=84+21\sqrt{15}-24-6\sqrt{15}-15\sqrt{15}\)
=60
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7a) \(\Delta=\left(3m+1\right)^2-4\left(2m^2+m-1\right)=m^2+2m+5=\left(m+1\right)^2+4>0\)
\(\Rightarrow\) pt luôn có 2 nghiệm phân biệt
b) Áp dụng hệ thức Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=3m+1\\x_1x_2=2m^2+m-1\end{matrix}\right.\)
Ta có: \(x_1^2+x_2^2-3x_1x_2=\left(x_1+x_2\right)^2-5x_1x_2=\left(3m+1\right)^2-5\left(2m^2+m-1\right)\)
\(=-m^2+m+6=-\left(m^2-m-6\right)\)
Ta có: \(m^2-m-6=m^2-2.m.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{25}{4}\)
\(=\left(m-\dfrac{1}{2}\right)^2-\dfrac{25}{4}\ge-\dfrac{25}{4}\Rightarrow-\left(m^2-m-6\right)\le\dfrac{25}{4}\)
\(\Rightarrow GTLN=\dfrac{25}{4}\) khi \(m=\dfrac{1}{2}\)
a) Ta có: \(x^2-\left(3m+1\right)x+2m^2+m-1\)
\(\Delta=\left(3m+1\right)^2-4\left(2m^2+m-1\right)\)
\(=9m^2+6m+1-8m^2-4m+4\)
\(=m^2+2m+5\)
\(=\left(m+1\right)^2+4>0\forall m\)
Do đó: Phương trình luôn có hai nghiệm phân biệt với mọi m
b) Áp dụng hệ thức Vi-et, ta được:
\(\left\{{}\begin{matrix}x_1+x_2=3m+1\\x_1x_2=2m^2+m-1\end{matrix}\right.\)
Ta có: \(B=x_1^2+x_2^2-3x_1x_2\)
\(=\left(x_1+x_2\right)^2-5x_1x_2\)
\(=\left(3m+1\right)^2-5\left(2m^2+m-1\right)\)
\(=9m^2+6m+1-10m^2-5m+5\)
\(=-m^2+m+6\)
\(=-\left(m^2-m-6\right)\)
\(=-\left(m^2-2\cdot m\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{25}{4}\)
\(=-\left(m-\dfrac{1}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall m\)
Dấu '=' xảy ra khi \(m=\dfrac{1}{2}\)
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Cho mình hỏi : A = ( x thuộc N / 2x + 2 ; x bé hơn 100 mình cần gấp lắm rồi,
8:
a: Để đây là hsbn thì m-1<>0
=>m<>1
b: Để hàm số đồng biến thì m-1>0
=>m>1
Để hàm số nghịch biến thì m-1<0
=>m<1
c: Thay x=3 và y=4 vào (d) ta được:
3(m-1)+5=4
=>3m+2=4
=>3m=2
=>m=2/3
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x^2 + x - 2 = 0
<=> ( x^2 - x ) + ( 2x - 2 ) = 0
<=> x . ( x - 1 ) + 2 . ( x - 1 ) = 0
<=> ( x - 1 ) . ( x + 2 ) = 0
<=> x - 1 = 0 hoặc x + 2 = 0
<=> x = 1 hoặc x = -2
Vậy .......
Tk mk nha
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a: \(P=\dfrac{2\sqrt{x}-9-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-x+2\sqrt{x}+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
b: Để P<1 thì P-1<0
\(\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\)
\(\Leftrightarrow\sqrt{x}-3< 0\)
hay x<9
Kết hợp ĐKXĐ, ta được: 0<=x<9 và x<>4
c: Để P<1 thì 0<=x<9 và x<>4
mà x là số nguyên
nên \(x\in\left\{0;1;2;3;5;6;7;8\right\}\)