thu gọn đa thức :
a, \(\left(2x-3\right)\left(2x+3\right)\)
b,\(\left(\sqrt{5}-x\right)\left(\sqrt{5}+x\right)\)
giải pt :
a, \(\left(2x-6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
Mình rút gọn như thế này đúng không nhỉ?
\(P=\left(2-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(P=\left[\frac{2\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right]:\left[\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right]\)
\(P=\left(\frac{4\sqrt{x}-6}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\left(\frac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}:\frac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}.\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}{2x+3\sqrt{x}+1}\)
\(P=\left(3\sqrt{x}-5\right).\frac{\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x+3\sqrt{x}-5\sqrt{x}-5}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x-5\sqrt{x}-5}{2x+1}\)
từ dòng cuối là sai rồi bạn à
Bạn bỏ dòng cuối đi còn lại đúng rồi
Ở tử đặt nhân tử chung căn x chung rồi lại đặt căn x +1 chung
Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra
rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
giải pt :a,\(\left(2x+6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
Tính DKXD của các căn bậc thức sau:
a)\(\sqrt{2x-4}\)
b)\(\sqrt{\dfrac{3}{-2x+1}}\)
c)\(\sqrt{\dfrac{-3x+5}{-4}}\)
d)\(\sqrt{-5\left(-2x+6\right)}\)
e)\(\sqrt{\left(x^2+2\right)\left(x-3\right)}\)
f)\(\sqrt{\dfrac{x^2+5}{-x+2}}\)
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 ...
giải pt :
a,\(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
b, \(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
Thu gọn đa thức
\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)\)
= 4x2-4x+1+x2+3x+9-5(x2-49)
=5x2-x+10-5x2+245
=-x+255
a. \(\sqrt{\left(2x+3\right)^2}=x+1\)
b. \(\sqrt{\left(2x-1\right)^2}=x+1\)
c. \(\sqrt{x+3}=5\)
d. \(\sqrt{x+2}=\sqrt{7}\)
e. \(5\sqrt{x}=20\)
f. \(\sqrt{x+4}=7\)
g. \(\sqrt{\left(2x+1\right)^2}=3\)
a, \(\sqrt{\left(2x+3\right)^2}=x+1\)
\(\Leftrightarrow\left|2x+3\right|=x+1\)
TH1: \(\left\{{}\begin{matrix}2x+3=x+1\\2x+3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x\ge-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\) vô nghiệm.
Vậy phương trình vô nghiệm.
TH2: \(\left\{{}\begin{matrix}-2x-3=x+1\\2x+3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{4}{3}\\x< -\dfrac{3}{2}\end{matrix}\right.\Rightarrow\) vô nghiệm.
b,
a, \(\sqrt{\left(2x-1\right)^2}=x+1\)
\(\Leftrightarrow\left|2x-1\right|=x+1\)
TH1: \(\left\{{}\begin{matrix}2x-1=x+1\\2x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x\ge\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x=2\)
TH2: \(\left\{{}\begin{matrix}-2x+1=x+1\\2x-1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x< \dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x=0\)
giải pt:
a,\(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
b,\(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
1, Rút gọn biểu thức: \(A=\dfrac{-3}{4}.\sqrt{9-4\sqrt{5}}.\sqrt{\left(-8\right)^2.\left(2+\sqrt{5}\right)^2}\)
2, Với \(x=\sqrt{4+2\sqrt{3}}\). Tính giá trị biểu thức \(P=x^2-2x+2020\)
Bài 2:
\(x=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
Ta có: \(P=x^2-2x+2020\)
\(=4+2\sqrt{3}-2\left(\sqrt{3}-1\right)+2020\)
\(=4+2\sqrt{3}-2\sqrt{3}+2+2020\)
=2026
Bài 1:
\(A=-\dfrac{3}{4}\cdot\sqrt{9-4\sqrt{5}}\cdot\sqrt{\left(-8\right)^2\cdot\left(2+\sqrt{5}\right)^2}\)
\(=\dfrac{-3}{4}\cdot8\cdot\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)\)
=-6