\(\sqrt{x}\)

10) a + 3\(\sqrt{a}\)- 10

11)...">

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29 tháng 8 2021

\(9,x-\sqrt{x}=\sqrt{x}\left(\sqrt{x}-1\right)\\ 10,a+3\sqrt{a}-10=\left(\sqrt{a}-1\right)\left(\sqrt{a}+5\right)\\ 11,x+\sqrt{x}-2=\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\\ 12,x-5\sqrt{x}+6=\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)\)

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

10: \(a+3\sqrt{a}-10=\left(\sqrt{a}+5\right)\left(\sqrt{a}-2\right)\)

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

12: \(x-5\sqrt{x}+6=\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)\)

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

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

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

\(=3\left(x+2\cdot\sqrt{x}\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{9}\right)\)

\(=3\left(\sqrt{x}+\dfrac{1}{3}\right)^2+\dfrac{2}{3}>=3\cdot\dfrac{1}{9}+\dfrac{2}{3}=1\)

Dấu '=' xảy ra khi x=0

2: \(=x+3\sqrt{x}+\dfrac{9}{4}-\dfrac{21}{4}=\left(\sqrt{x}+\dfrac{3}{2}\right)^2-\dfrac{21}{4}>=-3\)

Dấu '=' xảy ra khi x=0

3: \(A=-2x-3\sqrt{x}+2< =2\)

Dấu '=' xảy ra khi x=0

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

Dấu '=' xảy ra khi x=1

13 tháng 6 2018

Mình làm một vài câu thôi nhé, các câu còn lại tương tự.

Giải:

a) ??? Đề thiếu

b) \(\sqrt{-3x+4}=12\)

\(\Leftrightarrow-3x+4=144\)

\(\Leftrightarrow-3x=140\)

\(\Leftrightarrow x=\dfrac{-140}{3}\)

Vậy ...

c), d), g), h), i), p), q), v), a') Tương tự b)

w), x) Mình đã làm ở đây:

Câu hỏi của Ami Yên - Toán lớp 9 | Học trực tuyến

z) \(\sqrt{16\left(x+1\right)^2}-\sqrt{9\left(x+1\right)^2}=4\)

\(\Leftrightarrow4\left(x+1\right)-3\left(x+1\right)=4\)

\(\Leftrightarrow x+1=4\)

\(\Leftrightarrow x=3\)

Vậy ...

b') \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)

\(\Leftrightarrow3\sqrt{x+1}+2\sqrt{x+1}=\sqrt{x+1}\)

\(\Leftrightarrow3\sqrt{x+1}+2\sqrt{x+1}-\sqrt{x+1}=0\)

\(\Leftrightarrow4\sqrt{x+1}=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy ...

13 tháng 6 2018

- Câu a có chút thiếu sót, mong thông cảm :)

\(\sqrt{3x-1}\) = 4

1 tháng 8 2018

\(a\text{) }\sqrt{10+\sqrt{9}}=\sqrt{10+3}=\sqrt{13}\)

\(b\text{) }\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}\\ =\sqrt{18+3+2\sqrt{54}}-\sqrt{18+3-2\sqrt{54}}\\ =\sqrt{\left(\sqrt{18}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}\\ =\sqrt{18}+\sqrt{3}-\sqrt{18}+\sqrt{3}\\ =2\sqrt{3}\)

\(d\text{) }\sqrt{x+1+2\sqrt{x}}\left(x\ge0\right)\\ =\sqrt{\left(\sqrt{x}+1\right)^2}=\sqrt{x}+1\)

\(e\text{) }\sqrt{2x+3+2\sqrt{x^2+3x+2}}\left(x\le-2;x\ge-1\right)\\ =\sqrt{\left(x+2\right)+\left(x+1\right)+2\sqrt{\left(x+1\right)\left(x+2\right)}}=\sqrt{\left(\sqrt{x+1}+\sqrt{x+2}\right)^2}=\sqrt{x+1}+\sqrt{x+2}\)

Xem lại đề câu c nha.

1 tháng 8 2018

a)\(\sqrt{10+\sqrt{9}}=\sqrt{10+3}=\sqrt{13}\)

b)\(\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}\)

=\(\sqrt{\left(3\sqrt{2}\right)^2+2.3\sqrt{2}.\sqrt{3}+\sqrt{3^2}}-\sqrt{\left(3\sqrt{2}\right)^2-2.3.\sqrt{2}.\sqrt{3}+\sqrt{3^2}}\)

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

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

=\(2\sqrt{3}\)

c)\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10-2\sqrt{5}}}\)

ÁP dụng HĐT \(\sqrt{a+b}\pm\sqrt{a-b}=\sqrt{2\left(a.\sqrt{a^2\pm b}\right)}\)ta có:

=\(\sqrt{2\left(4+\sqrt{4^2-10-2\sqrt{5}}\right)}\)

=\(\sqrt{2\left(4+\sqrt{16-10-2\sqrt{5}}\right)}\)

=\(\sqrt{2\left(4+\sqrt{6-2\sqrt{5}}\right)}\)

=\(\sqrt{2\left(4+\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}.1+1^2}\right)}\)

=\(\sqrt{2\left(4+\sqrt{\left(\sqrt{5}-1\right)^2}\right)}\)

=\(\sqrt{2\left(4+\sqrt{5}-1\right)}\)

=\(\sqrt{2\left(3+\sqrt{5}\right)}\)

=\(\sqrt{6+\sqrt{5}}=\sqrt{5}+1\)

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

2 tháng 10 2019

mầy câu 1;3;;4;5 cách làm nhu nhau(nhân liên hop hoac bình phuong lên)

1.

\(DK:x\in\left[-4;5\right]\)

\(\Leftrightarrow\sqrt{x-5}+\left(\sqrt{x+4}-3\right)=0\)

\(\Leftrightarrow\sqrt{x-5}+\frac{x-5}{\sqrt{x+4}+3}=0\)

\(\Leftrightarrow\sqrt{x-5}\left(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}\right)=0\)

Vi \(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}>0\)

\(\Rightarrow\sqrt{x-5}=0\)

\(x=5\left(n\right)\)

Vay nghiem cua PT la \(x=5\)

2 tháng 10 2019

2.

\(DK:x\ge0\)

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

\(\Leftrightarrow|\sqrt{x}-2|+|\sqrt{x}-3|=1\)

Ta co:

\(|\sqrt{x}-2|+|\sqrt{x}-3|=|\sqrt{x}-2|+|3-\sqrt{x}|\ge|\sqrt{x}-2+3-\sqrt{x}|=1\)

Dau '=' xay ra khi \(\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)\ge0\)

TH1:

\(\hept{\begin{cases}\sqrt{x}-2\ge0\\3-\sqrt{x}\ge0\end{cases}\Leftrightarrow4\le x\le9\left(n\right)}\)

TH2:(loai)

Vay nghiem cua PT la \(x\in\left[4;9\right]\)

NV
7 tháng 10 2019

Bạn tự tìm ĐKXĐ.

a/ \(\sqrt{4-5x}=12\Rightarrow4-5x=144\Rightarrow x=-28\)

b/ \(10+\sqrt{3x}=\left(2+\sqrt{6}\right)^2=10+4\sqrt{6}\)

\(\Rightarrow\sqrt{3x}=4\sqrt{6}\Rightarrow\sqrt{x}=4\sqrt{2}\)

\(\Rightarrow x=32\)

c/ \(2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Rightarrow\sqrt{x+5}=2\Rightarrow x+5=4\Rightarrow x=-1\)

d/ \(\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)

\(\Rightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=6\end{matrix}\right.\)

e/ \(\sqrt{\frac{4x+3}{x+1}}=3\Leftrightarrow\frac{4x+3}{x+1}=9\)

\(\Rightarrow4x+3=9x+9\Rightarrow5x=-6\Rightarrow x=-\frac{6}{5}\)

f/ \(\sqrt{x-2}\le3\Rightarrow x-2\le9\Rightarrow2\le x\le11\)