Rút gọn biểu thức x 2 - 6 x + 9 x - 3 với x < 3 ta được:
A. −1
B. 1
C. 2
D. −2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
19 tháng 1 2022
Bài 1:
a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để A=3 thì 3x-9=x+1
=>2x=10
hay x=5
Bài 2:
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)
b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
2
18 tháng 4 2023
`a)|x-2|=2<=>[(x=4(ko t//m)),(x=0(t//m)):}`
Thay `x=0` vào `A` có: `A=[2\sqrt{0}-3]/[\sqrt{0}-2]=3/2`
`b)` Với `x >= 0,x ne 4` có:
`B=[2(\sqrt{x}-3)+\sqrt{x}(\sqrt{x}+3)-4\sqrt{x}]/[(\sqrt{x}+3)(\sqrt{x}-3)]`
`B=[2\sqrt{x}-6+x+3\sqrt{x}-4\sqrt{x}]/[(\sqrt{x}+3)(\sqrt{x}-3)]`
`B=[x+\sqrt{x}-6]/[(\sqrt{x}+3)(\sqrt{x}-3)]`
`B=[(\sqrt{x}+3)(\sqrt{x}-2)]/[(\sqrt{x}+3)(\sqrt{x}-3)]`
`B=[\sqrt{x}-2]/[\sqrt{x}-3]`
`c)` Với `x >= 0,x ne 4` có:
`C=A.B=[2\sqrt{x}-3]/[\sqrt{x}-2].[\sqrt{x}-2]/[\sqrt{x}-3]=[2\sqrt{x}-3]/[\sqrt{x}-3]`
Có: `C >= 1`
`<=>[2\sqrt{x}-3]/[\sqrt{x}-3] >= 1`
`<=>[2\sqrt{x}-3-\sqrt{x}+3]/[\sqrt{x}-3] >= 0`
`<=>[\sqrt{x}]/[\sqrt{x}-3] >= 0`
Vì `x >= 0=>\sqrt{x} >= 0`
`=>\sqrt{x}-3 > 0`
`<=>x > 9` (t/m đk)
26 tháng 7 2023
a: A=x+3+|x-3|
=x+3+3-x(x<=3)
=6
b:\(B=\sqrt{x^2+4x+4}-\sqrt{x^2}\)
\(=\left|x+2\right|-\left|x\right|\)
=x+2-x=2
c: \(C=\dfrac{\sqrt{x^2-2x+1}}{x-1}\)
\(=\dfrac{\left|x-1\right|}{x-1}=\dfrac{x-1}{x-1}=1\)
3 tháng 12 2023
a: Sửa đề: \(B=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
Khi x=9 thì \(B=\dfrac{\sqrt{9}+1}{\sqrt{9}+2}\)
\(=\dfrac{3+1}{3+2}=\dfrac{4}{5}\)
b: \(A=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{6+\sqrt{x}}{x-4}\)
\(=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}+6}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)+\sqrt{x}\left(\sqrt{x}+2\right)-\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-5\sqrt{x}+6+x+2\sqrt{x}-\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2x-4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{2\sqrt{x}}{\sqrt{x}+2}\)
c: P=A/B
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+2}:\dfrac{\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)
\(P-2=\dfrac{2\sqrt{x}}{\sqrt{x}+1}-2=\dfrac{2\sqrt{x}-2\sqrt{x}-2}{\sqrt{x}+1}\)
\(=\dfrac{-2}{\sqrt{x}+1}< 0\)
=>P<2
22 tháng 10 2023
1:
a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)
\(=4x^2-20x+25-4x^2-12x\)
=-32x+25
b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)
\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)
c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)
\(=\left(x-1-x-2\right)^2+5\left(2x-3\right)\)
\(=\left(-3\right)^2+5\left(2x-3\right)\)
\(=9+10x-15=10x-6\)
2:
a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)
\(=9x^2-12x+4-5x^2+20x+4x-4\)
\(=4x^2+12x\)
b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)
\(=27-x^3+x^3-9x^2+27x-27\)
\(=-9x^2+27x\)
c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)
\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)
\(=-5\left(x^2-16\right)=-5x^2+80\)
22 tháng 7 2023
\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{x-9}\right)\cdot\dfrac{2\sqrt{x}+6}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}-3+2\sqrt{x}}{x-9}\cdot\dfrac{2\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\cdot\dfrac{2}{\sqrt{x}-3}=\dfrac{6}{\sqrt{x}-3}\)
25 tháng 7 2021
a) Ta có: \(\left(x-2\right)^3-\left(3+x^2\right)\left(3-x\right)\)
\(=x^3-6x^2+12x-8+\left(x-3\right)\left(x^2+3\right)\)
\(=x^3-6x^2+12x-8+x^3+3x-3x^2-9\)
\(=2x^3-9x^2+15x-17\)
b) Ta có: \(x\left(x-14\right)-10\left(x-1\right)^2\)
\(=x^2-14x-10\left(x^2-2x+1\right)\)
\(=x^2-14x-10x^2+20x-10\)
\(=-9x^2+6x-10\)
c) Ta có: \(2x\left(x+2\right)-\left(x+2\right)\left(x-2\right)\)
\(=2x^2+4x-\left(x^2-4\right)\)
\(=2x^2+4x-x^2+4\)
\(=x^2+4x+4\)
d) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-\left(x^3-27\right)\)
\(=x^3-27-x^3+27\)
=0
Đáp án đúng : A