\(2M=\left(\sqrt{x^2-3x+25}-\sqrt{x^2-3x+9}\right)\)\(\left(\sqrt{x^2-3x+25}+\sqrt{x^2-3x+9}\right)\)

\(2M=x^2-3x+25-x^2+3x-9=16\)

M = 8

a: TH1: x>=2

A=x+x-2=2x-2

TH2: x<2

A=x+2-x=2

b: TH1: x>=3

A=x-3-x=-3

TH2: x<3

A=3-x-x=-2x+3

c: TH1: x>=1

C=x-x+1=1

TH2: x<1

C=x+x-1=2x-1

d: TH1: m>=3

C=m-3-2m=-3-m

TH2: m<3

C=-m+3-2m=-3m+3

e: TH1: m>=1

E=m-m+1=1

TH2: m<1

E=m+m-1=2m-1

Thiếu soát gì mog bạn thông cảm :]

6) \(\sqrt{x^2-4x+1}=x\left(x\ge0\right)\) 

\(\Leftrightarrow x^2-4x+1=x^2\)

\(\Leftrightarrow x^2-x^2=4x-1\)

\(\Leftrightarrow4x=1\)

\(\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\) 

8) \(\sqrt{x^2-x-6}=\sqrt{x-3}\left(x\ge3\right)\) 

\(\Leftrightarrow x^2-x-6=x-3\)

\(\Leftrightarrow x^2-2x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

9) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\left(x\ge1\right)\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

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

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

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=1+1\)

\(\Leftrightarrow x=2\left(tm\right)\)

Câu 1:

\(A=\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{x+9\sqrt{x}}{x-9}\left(x\ge0;x\ne9\right)\)

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

\(=\dfrac{2x+6\sqrt{x}-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)\(=\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

Câu 2:

\(V\left(3\right)=12000000-1400000.3=7800000\)

Có: \(V\left(t\right)=6400000\) \(\Leftrightarrow12000000-1400000t=6400000\)

\(\Leftrightarrow t=4\) => Sau 4 năm thì gtri chiếc máy tính này còn 6400000 đ

b,\(\left\{{}\begin{matrix}2x+y=5\\mx+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{4-mx}{3}=5\\y=\dfrac{4-mx}{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\left(6-m\right)=11\left(1\right)\\y=\dfrac{4-mx}{3}\end{matrix}\right.\)

Xét \(m=6\) thay vào pt ta đc \(\left\{{}\begin{matrix}2x+y=5\\6x+3y=4\end{matrix}\right.\) (vô nghiệm)

\(\Rightarrow m\ne6\)

Từ (1) \(\Rightarrow x=\dfrac{11}{6-m}\)

 \(\Rightarrow y=\dfrac{4-\dfrac{11m}{6-m}}{3}\)\(=\dfrac{24-15m}{3\left(6-m\right)}\)

\(xy>0\Leftrightarrow\dfrac{11}{6-m}.\dfrac{24-15m}{3\left(6-m\right)}>0\)

\(\Leftrightarrow\dfrac{11\left(24-15m\right)}{3\left(6-m\right)^2}>0\) 

\(\Leftrightarrow24-15m>0\Leftrightarrow m< \dfrac{24}{15}\)

`A=(2sqrtx)/(sqrtx-3)-(x+9sqrtx)/(x-9)`
`đk:x>=0,x ne 9`
`A=(2x+6sqrtx)/(x-9)-(x+9sqrtx)/(x-9)`
`=(x-3sqrtx)/(x-9)`
`=sqrtx/(sqrtx+3)`

ĐKXĐ: -3 ≤ x ≤ 3

Ta có √25 − x^2 − √9 − x^2 = 2

<=> √25 − x^2 = 2 + √9 − x^2

<=> (√25 − x^2)^2 = (2 + √9 − x^2)^2

<=> 25 - x^2 = 4 + 4.√9 − x^2 + 9 - x^2

<=> 12 - x^2 = 4.√9 − x^2 - x^2

<=> 12 = 4.√9 − x^2

<=> √9 − x^2 = 3

<=> 9 − x^2 = 9 <=> x^2 = 0 <=> x = 0

Vậy x = 0

a. ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{3}{2}.\sqrt{9}.\sqrt{x-1}+24.\sqrt{\frac{1}{64}}.\sqrt{x-1}=-17$

$\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17$

$\Leftrightarrow -\sqrt{x-1}=-17$

$\Leftrightarrow \sqrt{x-1}=17$

$\Leftrightarrow x-1=289$

$\Leftrightarrow x=290$

b. ĐKXĐ: $x\geq \frac{1}{2}$

PT $\Leftrightarrow \sqrt{9}.\sqrt{2x-1}-0,5\sqrt{2x-1}+\frac{1}{2}.\sqrt{25}.\sqrt{2x-1}+\sqrt{49}.\sqrt{2x-1}=24$

$\Leftrightarrow 3\sqrt{2x-1}-0,5\sqrt{2x-1}+2,5\sqrt{2x-1}+7\sqrt{2x-1}=24$
$\Leftrightarrow 12\sqrt{2x-1}=24$

$\Leftrihgtarrow \sqrt{2x-1}=2$

$\Leftrightarrow x=2,5$ (tm)

c. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{36}.\sqrt{x-2}-15\sqrt{\frac{1}{25}}\sqrt{x-2}=4(5+\sqrt{x-2})$

$\Leftrightarrow 6\sqrt{x-2}-3\sqrt{x-2}=20+4\sqrt{x-2}$

$\Leftrightarrow \sqrt{x-2}=-20< 0$ (vô lý)

Vậy pt vô nghiệm

d. ĐKXĐ: $x>\frac{-2}{3}$

PT $\Leftrightarrow \sqrt{\frac{1}{3x+2}}-\frac{1}{2}\sqrt{9}.\sqrt{\frac{1}{3x+2}}+\sqrt{16}.\sqrt{\frac{1}{3x+2}}-5\sqrt{\frac{1}{4}}\sqrt{\frac{1}{3x+2}}=1$

$\Leftrightarrow \sqrt{\frac{1}{3x+2}}-\frac{3}{2}\sqrt{\frac{1}{3x+2}}+4\sqrt{\frac{1}{3x+2}}-\frac{5}{2}\sqrt{\frac{1}{3x+2}}=1$

$\Leftrightarrow \sqrt{\frac{1}{3x+2}}=1$

$\Leftrightarrow \frac{1}{3x+2}=1$

$\Leftrightarrow 3x+2=1$

$\Leftrightarrow x=-\frac{1}{3}$

\(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)\)

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

\(\dfrac{x-25}{\sqrt{x}-5}-\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\sqrt{x}-5}-\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}+2}\)

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

a: Ta có: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}+\sqrt{2}\right)\)

\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)

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

=3-1

=2

b: Ta có: \(\dfrac{x-25}{\sqrt{x}-5}-\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}\)

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

=3

a) Đk: \(x>0;x\ne9;x\ne25\)

Đặt \(A=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

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

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

\(=\dfrac{3\sqrt{x}+x}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{-\sqrt{x}+5}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\sqrt{x}\left(3+\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\sqrt{x}-5}\)

\(=\dfrac{x}{\sqrt{x}-5}\)

b) Đk: \(x\ge0;x\ne1;x\ne25\)

Biểu thức

\(=\left[\dfrac{\sqrt{x}-2}{\sqrt{x}+5}+\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{x+9}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right]:\dfrac{1-\sqrt{x}}{5+\sqrt{x}}\)

\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)+\sqrt{x}\left(\sqrt{x}+5\right)-x-9}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}+5}{1-\sqrt{x}}\)

\(=\dfrac{x-7\sqrt{x}+10+x+5\sqrt{x}-x-9}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}\)

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

Đk: \(x>0;x\ne4\)

Biểu thức:

\(=\left[\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{1}{\left(\sqrt{x}-2\right)^2}\right].\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}}\)

\(=\left[\dfrac{-1}{\left(2-\sqrt{x}\right)\left(\sqrt{x}+2\right)}-\dfrac{1}{\left(2-\sqrt{x}\right)^2}\right].\left(2-\sqrt{x}\right)\)

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

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

Khoảng cách hai ta là 100 bước, em bước 100, anh lùi 1