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![](https://rs.olm.vn/images/avt/0.png?1311)
a)đk:`2x-4>=0`
`<=>2x>=4`
`<=>x>=2.`
b)đk:`3/(-2x+1)>=0`
Mà `3>0`
`=>-2x+1>=0`
`<=>1>=2x`
`<=>x<=1/2`
c)`đk:(-3x+5)/(-4)>=0`
`<=>(3x-5)/4>=0`
`<=>3x-5>=0`
`<=>3x>=5`
`<=>x>=5/3`
d)`đk:-5(-2x+6)>=0`
`<=>-2x+6<=0`
`<=>2x-6>=0`
`<=>2x>=6`
`<=>x>=3`
e)`đk:(x^2+2)(x-3)>=0`
Mà `x^2+2>=2>0`
`<=>x-3>=0`
`<=>x>=3`
f)`đk:(x^2+5)/(-x+2)>=0`
Mà `x^2+5>=5>0`
`<=>-x+2>0`
`<=>-x>=-2`
`<=>x<=2`
a, ĐKXĐ : \(2x-4\ge0\)
\(\Leftrightarrow x\ge\dfrac{4}{2}=2\)
Vậy ..
b, ĐKXĐ : \(\left\{{}\begin{matrix}\dfrac{3}{-2x+1}\ge0\\-2x+1\ne0\end{matrix}\right.\)
\(\Leftrightarrow-2x+1>0\)
\(\Leftrightarrow x< \dfrac{1}{2}\)
Vậy ..
c, ĐKXĐ : \(\dfrac{-3x+5}{-4}\ge0\)
\(\Leftrightarrow-3x+5\le0\)
\(\Leftrightarrow x\ge\dfrac{5}{3}\)
Vậy ...
d, ĐKXĐ : \(-5\left(-2x+6\right)\ge0\)
\(\Leftrightarrow-2x+6\le0\)
\(\Leftrightarrow x\ge-\dfrac{6}{-2}=3\)
Vậy ...
e, ĐKXĐ : \(\left(x^2+2\right)\left(x-3\right)\ge0\)
\(\Leftrightarrow x-3\ge0\)
\(\Leftrightarrow x\ge3\)
Vậy ...
f, ĐKXĐ : \(\left\{{}\begin{matrix}\dfrac{x^2+5}{-x+2}\ge0\\-x+2\ne0\end{matrix}\right.\)
\(\Leftrightarrow-x+2>0\)
\(\Leftrightarrow x< 2\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
1. b) \(\left(x\sqrt{\dfrac{6}{x}}+\sqrt{\dfrac{2x}{3}}+\sqrt{6x}\right):\sqrt{6x}\)
=\(\left(x\sqrt{\dfrac{6x}{x^2}}+\sqrt{\dfrac{6x}{9}}+\sqrt{6x}\right):\sqrt{6x}\)
=\(\left(\sqrt{6x}+\dfrac{1}{3}\sqrt{6x}+\sqrt{6x}\right):\sqrt{6x}\)
=\(\dfrac{7}{3}\sqrt{6x}:\sqrt{6x}=\dfrac{7}{3}\)
2.
P=\(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)(bn có ghi sai đề ko)
a) ĐKXĐ : \(x\ge1,x\ge2,x\ge0\)
b) P=\(\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}+\dfrac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
=\(\dfrac{x-3\sqrt{x}-\sqrt{x}+3-2x+\sqrt{x}+4\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
=\(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\dfrac{1}{\sqrt{x}-2}\)
c) thay x= \(4-2\sqrt{3}\)vào P ta có :
\(\dfrac{1}{\sqrt{4-2\sqrt{3}}-2}=\dfrac{1}{\sqrt{3}-1-2}=\dfrac{1}{\sqrt{3}-3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
\(=\sqrt{8-4\sqrt{3}}\cdot\sqrt{\dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}}\)
\(=\sqrt{8-4\sqrt{3}}\cdot\sqrt{\dfrac{\left(\sqrt{6}+\sqrt{2}\right)^2}{6-2}}\)
\(=\left(\sqrt{6}-\sqrt{2}\right)\cdot\dfrac{\left(\sqrt{6}+\sqrt{2}\right)}{2}\)
\(=\dfrac{6-2}{2}=\dfrac{4}{2}=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(A=\left|2a-1\right|-2a\)
TH1: a>=1/2
A=2a-1-2a=-1
TH2: a<1/2
A=1-2a-2a=1-4a
b: \(B=x-2y-\left|x-2y\right|\)
TH1: x>=2y
A=x-2y-x+2y=0
TH2: x<2y
A=x-2y+x-2y=2x-4y
c: \(=x^2+\left|x^2-4\right|\)
TH1: x>=2 hoặc x<=-2
\(A=x^2+x^2-4=2x^2-4\)
TH2: -2<x<2
\(A=x^2+4-x^2=4\)
d: \(D=2x-1-\dfrac{\left|x-5\right|}{x-5}\)
TH1: x>5
\(D=2x-1-1=2x-2\)
TH2: x<5
D=2x-1+1=2x
![](https://rs.olm.vn/images/avt/0.png?1311)
a, P=\(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{\left(1-x\right)^2}{2}\)
\(P=\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\dfrac{\left(1-x\right)^2}{2}\)
\(P=\dfrac{x-\sqrt{x}-2-\left(x+\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\left(1-x\right)^2}{2}\)
\(P=\dfrac{-2\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(x-1\right)^2}{2}\)
\(P=\dfrac{-\sqrt{x}\left(x-1\right)}{\sqrt{x}+1}=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=-\sqrt{x}\left(\sqrt{x}-1\right)=\sqrt{x}-x\)b,x=\(7-4\sqrt{3}=4-2.2\sqrt{3}+3=\left(2-\sqrt{3}\right)^2\)
Thay vào ta có \(P=\sqrt{\left(4-\sqrt{3}\right)^2}-\left(7-4\sqrt{3}\right)\)
\(P=\left|4-\sqrt{3}\right|-7-4\sqrt{3}=4-\sqrt{3}-7+4\sqrt{3}\)
\(P=-3+3\sqrt{3}\)
Câu 2:
a: f(1)=2
=>m-1+2m-3=2
=>3m=6
=>m=2
=>f(x)=x+1
=>f(2)=2+1=3
b: f(-3)=0
=>-3m+3+2m-3=0
=>m=0
=>f(x)=-x-3
=>f(x) nghịch biến
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
a)
\(y=f\left(x\right)=2x^2\) | -5 | -3 | 0 | 3 | 5 |
f(x) | 50 | 18 | 0 | 18 | 50 |
b) Ta có: f(x)=8
\(\Leftrightarrow2x^2=8\)
\(\Leftrightarrow x^2=4\)
hay \(x\in\left\{2;-2\right\}\)
Vậy: Để f(x)=8 thì \(x\in\left\{2;-2\right\}\)
Ta có: \(f\left(x\right)=6-4\sqrt{2}\)
\(\Leftrightarrow2x^2=6-4\sqrt{2}\)
\(\Leftrightarrow x^2=3-2\sqrt{2}\)
\(\Leftrightarrow x=\sqrt{3-2\sqrt{2}}\)
hay \(x=\sqrt{2}-1\)
Vậy: Để \(f\left(x\right)=6-4\sqrt{2}\) thì \(x=\sqrt{2}-1\)
1. \(2M-N=\dfrac{2}{2-\sqrt{3}}-\sqrt{6}.\sqrt{2}=\dfrac{2-2\sqrt{3}\left(2-\sqrt{3}\right)}{2-\sqrt{3}}=\)\(\dfrac{2-4\sqrt{3}+6}{2-\sqrt{3}}=\dfrac{8-4\sqrt{3}}{2-\sqrt{3}}=4\)
Đáp án C
2. Ta có: A= \(-x+\sqrt{\left(6-x\right)^2}=-x+\left|6-x\right|\)
Mà x>6 \(\Rightarrow6-x< 0\)A=-x-6+x=-6
Đáp án C
3. Vẽ đồ thị hàm f(x) ta có:
Ta thấy f(2)<f(3), chọn Đáp án A
4.![](data:image/png;base64,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)
Khi đó, bán kính của đường tròn bằng \(\dfrac{2}{3}\)đường cao của tam giác đều ABC
Ta có: \(R=\dfrac{2}{3}.\dfrac{a\sqrt{3}}{2}=\dfrac{a\sqrt{3}}{3}\)
Đáp án A
Câu 1: C
Câu 2: C
Câu 3: A
Câu 4: A