\(\sqrt{2}\)                

b) 8 -...">

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26 tháng 8 2018

1) ta có : \(x\sqrt{x}+\sqrt{x}-x-1=\sqrt{x}\left(x+1\right)-\left(x+1\right)\)

\(=\left(\sqrt{x}-1\right)\left(x+1\right)\)

2) ta có : \(\sqrt{ab}-\sqrt{a}-\sqrt{b}+1=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)\)

\(=\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\)

3) ta có : \(x-\sqrt{x}-2=x+\sqrt{x}-2\sqrt{x}-2\)

\(=\sqrt{x}\left(\sqrt{x}+1\right)-2\left(\sqrt{x}+1\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\)

4) ta có : \(x-3\sqrt{x}+2=x-\sqrt{x}-2\sqrt{x}+2\)

\(=\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\)

5) ta có : \(-6x+5\sqrt{x}+1=-6x+6\sqrt{x}-\sqrt{x}+1\)

\(=6\sqrt{x}\left(1-\sqrt{x}\right)+\left(1-\sqrt{x}\right)=\left(6\sqrt{x}+1\right)\left(1-\sqrt{x}\right)\)

6) ta có : \(x+4\sqrt{x}+3=x+\sqrt{x}+3\sqrt{x}+3\)

\(=\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}+1\right)=\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)\)

7) ta có : \(3\sqrt{a}-2a-1=-2a+2\sqrt{a}+\sqrt{a}-1\)

\(=-2\sqrt{a}\left(\sqrt{a}-1\right)+\left(\sqrt{a}-1\right)=\left(1-2\sqrt{a}\right)\left(\sqrt{a}-1\right)\)

8) ta có : \(x+2\sqrt{x-1}=x-1+2\sqrt{x-1}+1\)

\(=\left(\sqrt{x-1}+1\right)^2\)

9) ta có : \(7\sqrt{x}-6x-2=-6x+3\sqrt{x}+4\sqrt{x}-2\)

\(=-3\sqrt{x}\left(2\sqrt{x}-1\right)+2\left(2\sqrt{x}-1\right)=\left(2-3\sqrt{x}\right)\left(2\sqrt{x}-1\right)\)

10) ta có : \(x-5\sqrt{x}+6=x-2\sqrt{x}-3\sqrt{x}+6\)

\(=\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)=\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)\)

11) ta có : \(x-2+\sqrt{x^2-4}=\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-2\right)\left(x+2\right)}\)

\(=\sqrt{x-2}\left(\sqrt{x-2}+\sqrt{x+2}\right)\)

16 tháng 7 2019

\(\text{a)}x\sqrt{x}+\sqrt{x}-x-1\)

\(=\left(x\sqrt{x}+\sqrt{x}\right)-\left(x+1\right)\)

\(=\sqrt{x}\left(x+1\right)-\left(x+1\right)\)

\(=\left(x+1\right)\left(\sqrt{x}-1\right)\)

\(\text{b)}\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6\)

\(=\left(\sqrt{ab}+2\sqrt{a}\right)+\left(3\sqrt{b}+6\right)\)

\(=\sqrt{a}\left(\sqrt{b}+2\right)+3\left(\sqrt{b}+2\right)\)

\(=\left(\sqrt{b}+2\right)\left(\sqrt{a}+3\right)\)

\(\text{c)}\left(1+\sqrt{x}\right)^2-4\sqrt{x}\)

\(=\left(1+\sqrt{x}\right)^2-\left(2\sqrt{\sqrt{x}}\right)^2\)

\(=\left(1+\sqrt{x}+2\sqrt{\sqrt{x}}\right)\left(1+\sqrt{x}-2\sqrt{\sqrt{x}}\right)\)

\(\text{d)}\sqrt{ab}-\sqrt{a}-\sqrt{b}+1\)

\(=\left(\sqrt{ab}-\sqrt{a}\right)-\left(\sqrt{b}-1\right)\)

\(=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)\)

\(=\left(\sqrt{b}-1\right)\left(\sqrt{a}-1\right)\)

\(\text{e)}a+\sqrt{a}+2\sqrt{ab}+2\sqrt{b}\)

\(=\left(a+\sqrt{a}\right)+\left(2\sqrt{ab}+2\sqrt{b}\right)\)

\(=\left[\left(\sqrt{a}\right)^2+\sqrt{a}\right]+\left(2\sqrt{ab}+2\sqrt{b}\right)\)

\(=\sqrt{a}\left(\sqrt{a}+1\right)+2\sqrt{b}\left(\sqrt{a}+1\right)\)

\(=\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\sqrt{b}\right)\)

\(\text{f)}x-2\sqrt{x-1}-a^2\)

\(=\left(\sqrt{x-2}\right)^2\left(\sqrt{\sqrt{x-1}}\right)^2-a^2\)

\(=\left(\sqrt{x-2}\sqrt{\sqrt{x-1}}\right)^2-a^2\)

\(=\left(\sqrt{x-2\sqrt{x-1}}\right)^2-a^2\)

\(=\left(\sqrt{x-2\sqrt{x-1}}+a\right)\left(\sqrt{x-2\sqrt{x-1}}-a\right)\)

Bài 1: Tìm x để căn thức sau có nghĩaa)\(\sqrt{x-3}\)    b) \(\sqrt{-3x}\)    c) \(\sqrt{\frac{5}{x+1}}\)    d) \(\sqrt{\frac{-10}{x^2+1}}\)Bài 2: Tínha) 3\(\sqrt{\left(-3\right)^2}\)    b) -5 \(\sqrt{\left(-2\right)^4}\)     c) \(\sqrt{\sqrt{\left(-10\right)^8}}\)    d) 2\(\sqrt{\left(-3\right)^4}\)\(+\)3\(\sqrt{\left(-2\right)^2}\)Bài 3: Rút gọna)\(\sqrt{\left(2+\sqrt{5}\right)^2}\)   b) \(\sqrt{\left(2-\sqrt{5}\right)^2}\)   c)...
Đọc tiếp

Bài 1: Tìm x để căn thức sau có nghĩa

a)\(\sqrt{x-3}\)    b) \(\sqrt{-3x}\)    c) \(\sqrt{\frac{5}{x+1}}\)    d) \(\sqrt{\frac{-10}{x^2+1}}\)

Bài 2: Tính

a) 3\(\sqrt{\left(-3\right)^2}\)    b) -5 \(\sqrt{\left(-2\right)^4}\)     c) \(\sqrt{\sqrt{\left(-10\right)^8}}\)    d) 2\(\sqrt{\left(-3\right)^4}\)\(+\)3\(\sqrt{\left(-2\right)^2}\)

Bài 3: Rút gọn

a)\(\sqrt{\left(2+\sqrt{5}\right)^2}\)   b) \(\sqrt{\left(2-\sqrt{5}\right)^2}\)   c) 2\(\sqrt{7}\)+\(\sqrt{\left(2-\sqrt{7}\right)^2}\) d) 3\(\sqrt{\left(x-5\right)^2}\) với x < 5

e)\(\sqrt{\frac{9+4\sqrt{5}}{\left(\sqrt{5+2}\right)^2}}\)     f)\(\sqrt{\frac{\sqrt{9-4\sqrt{5}}-\sqrt{5}}{2}}\)+ 5

Bài 4: Tìm x biết:

a)\(\sqrt{4x^2}\)= 8     b) \(\sqrt{1+4x+4x^2}\)\(=\)\(7\)    c)\(\sqrt{x^4}\)\(=\)\(3\)

Bài 5: Phân tích đa thức thành nhân tử

a) x2 -2      b) x2\(-\)2\(\sqrt{3}\)\(\times\)x \(+\)3

Bài 6: Chứng minh a\(\in\)z , b\(\in\)z

A=\(\sqrt{A-2\sqrt{5}}\)\(-\)\(\sqrt{6+2\sqrt{5}}\)   B=\(\frac{\sqrt{3-2\sqrt{2}}}{17-12\sqrt{2}}\)\(-\)\(\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)

1
5 tháng 8 2017

giúp mik vs thứ 2 mik nộp rr huhu

20 tháng 6 2021

a) 2a−4b=2(a−2b)2a−4b=2(a−2b)

c) 2ax−2ay+2a=2a(x−y+1)2ax−2ay+2a=2a(x−y+1)

e) 3xy(x−4)−9x(4−x)=3x(x−4)(y+3)3xy(x−4)−9x(4−x)=3x(x−4)(y+3)

b,d xem lại đề

20 tháng 6 2021

không hiểu

 what are you doing?

30 tháng 7 2018

\(\sqrt{16x^2}-2x=4x-2x=2x\)

a: \(=\sqrt{11}-1\)

b: \(=3\sqrt{3}+1\)

c: \(=\sqrt{3}+\sqrt{2}\)

d: \(=\sqrt{3}-\sqrt{2}\)

e: \(=\sqrt{3}-1\)

g: \(=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)