Rút gọn Biểu thức sau:
\(1+\sqrt{-1+9x^2}\)
Mn giúp m vs
K
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NT
2
13 tháng 10 2021
\(\left(3x-1\right)^2-9x\left(x+1\right)\)
\(=9x^2-6x+1-9x^2-9x\)
=-15x+1
TT
0
AN
1
24 tháng 10 2021
\(A=\dfrac{-4}{9x^2-4}+\dfrac{2x+1}{3x-2}-\dfrac{1}{3x+2}\)
\(=\dfrac{-4+6x^2+4x+3x+2-3x+2}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{6x^2+4x}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{2x}{3x-2}\)
19 tháng 11 2023
a: Sửa đề: \(M=3x-\sqrt[3]{27x^3+27x^2+9x+1}\)
\(=3x-\sqrt[3]{\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2+1^3}\)
\(=3x-\sqrt[3]{\left(3x+1\right)^3}\)
\(=3x-3x-1=-1\)
b: \(N=\sqrt[3]{8x^3+12x^2+6x+1}-\sqrt[3]{x^3}\)
\(=\sqrt[3]{\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2+1^3}-x\)
\(=\sqrt[3]{\left(2x+1\right)^3}-x\)
=2x+1-x
=x+1
1 tháng 11 2023
Sửa đề: \(A=\dfrac{1-3x}{2y}\cdot\sqrt{\dfrac{36y^2}{9x^2-6x+1}}\)
\(=\dfrac{1-3x}{2y}\cdot\sqrt{\left(\dfrac{6y}{3x-1}\right)^2}\)
\(=\dfrac{1-3x}{2y}\cdot\left|\dfrac{6y}{3x-1}\right|\)
x>1/3 nên 3x-1>0
y>0 nên 6y>0
=>\(A=\dfrac{1-3x}{2y}\cdot\dfrac{6y}{3x-1}=-3y\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 1
\(Q=\left(\dfrac{1}{2\sqrt{x}+1}+\dfrac{1}{2\sqrt{x}-1}\right):\dfrac{1}{1-4x}\)
\(=\left(\dfrac{2\sqrt{x}-1}{\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\dfrac{2\sqrt{x}+1}{\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\right).\left(1-4x\right)\)
\(=\left(\dfrac{2\sqrt{x}-1+2\sqrt{x}+1}{4x-1}\right)\left(1-4x\right)\)
\(=\dfrac{-4\sqrt{x}.\left(4x-1\right)}{4x-1}=-4\sqrt{x}\)
T
21 tháng 1
\(Q=\left(\dfrac{1}{2\sqrt{x}+1}+\dfrac{1}{2\sqrt{x}-1}\right):\dfrac{1}{1-4x}\left(dkxd:x\ge0;x\ne\dfrac{1}{4}\right)\)
\(=\left[\dfrac{2\sqrt{x}-1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}+1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\right]\cdot\left(1-4x\right)\)
\(=\dfrac{2\sqrt{x}-1+2\sqrt{x}+1}{4x-1}\cdot\left[-\left(4x-1\right)\right]\)
\(=4\sqrt{x}\cdot\left(-1\right)\)
\(=-4\sqrt{x}\)
13 tháng 6 2018
a, \(M=\sqrt{x^2-4x+4}-\sqrt{x^2+4x+4}\) (ĐK : \(\forall x\in R\))
\(=\sqrt{\left(x-2\right)^2}-\sqrt{\left(x+2\right)^2}\)
* Nếu x\(\ge2\Rightarrow M=x-2-x-2=-4\)
*Nếu x<2 => M=2-x-x-2=-2x
b,Để M=2\(\ne-4\)
=>M=-2x
=>-2x=-4
=>x=2
__________________________________________________________________________________________
P=\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
* Nếu \(x\ge2\Rightarrow P=\sqrt{x-1}+1+\sqrt{x-1}-1=2\sqrt{x-1}\)
* Nếu x<2 =>P=\(\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
VẬY.......
Tk nha!
NM
Nguyễn Minh Quang
Giáo viên
9 tháng 9 2021
\(\frac{3}{\sqrt{7}-1}+\frac{3}{\sqrt{7}+1}=\frac{3\left[\sqrt{7}+1+\sqrt{7}-1\right]}{\left(\sqrt{7}+1\right)\left(\sqrt{7}-1\right)}=\frac{6\sqrt{7}}{6}=\sqrt{7}\)
\(\frac{3}{\sqrt{X}-1}-\frac{2}{\sqrt{X}+1}+\frac{X-7}{X-1}=\frac{3\left(\sqrt{X}+1\right)-2\left(\sqrt{X}-1\right)+X-7}{\left(\sqrt{X}+1\right)\left(\sqrt{X}-1\right)}=\frac{X+\sqrt{X}-2}{\left(\sqrt{X}+1\right)\left(\sqrt{X}-1\right)}=\frac{\sqrt{X}+2}{\sqrt{X}+1}\)
VT
9 tháng 9 2021
TÍNH GIÁ TRỊ BIỂU THỨC:
\(\frac{3}{\sqrt{7}-1}\) + \(\frac{3}{\sqrt{7}+1}\)= \(\frac{3\left(\sqrt{7}+1\right)+3\left(\sqrt{7}-1\right)}{\left(\sqrt{7}-1\right)\left(\sqrt{7}+1\right)}\)= \(\frac{3\sqrt{7}+3+3\sqrt{7}-3}{6}\)=\(\frac{6\sqrt{7}}{6}\)=\(\sqrt{7}\)
RÚT GỌN BIỂU THỨC:
\(\frac{3}{\sqrt{X}-1}\)-\(\frac{2}{\sqrt{X}+1}\)+\(\frac{X-7}{X-1}\)
= \(\frac{3\left(\sqrt{X}+1\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)-\(\frac{2\left(\sqrt{X}-1\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)+\(\frac{X-7}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{3\sqrt{X}+3-2\sqrt{X}+2+X-7}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{X+\sqrt{X}-2}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{\left(\sqrt{X}+1\right)\left(\sqrt{X}-2\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{\sqrt{X}-2}{\sqrt{X}-1}\)
CHÚC EM HỌC TỐT!
16 tháng 7 2021
Ta có: M=A+B
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\)
\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)