a) ĐK: \(x\ge0\)

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

\(\Leftrightarrow2\sqrt{x}.\left(-\dfrac{1}{2}\right)+5=0\)

\(\Leftrightarrow x=25\) (thỏa)

Vậy \(x=25\)

b) Đk: \(x\le3\)

PT \(\Leftrightarrow\sqrt{3-x}-\sqrt{9\left(3-x\right)}+\dfrac{5}{4}\sqrt{16\left(3-x\right)}=6\)

\(\Leftrightarrow\sqrt{3-x}\left(1-\sqrt{9}+\dfrac{5}{4}.\sqrt{16}\right)=6\)

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

Vậy \(x=-1\)

2:

a: 

Sửa đề: \(P=\left(\dfrac{2}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\dfrac{2}{\sqrt{1-a^2}}+1\right)\)

\(P=\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}:\dfrac{2+\sqrt{1-a^2}}{\sqrt{1-a^2}}\)

\(=\dfrac{2+\sqrt{1-a^2}}{\sqrt{1+a}}\cdot\dfrac{\sqrt{1-a^2}}{2+\sqrt{1-a^2}}=\sqrt{\dfrac{1-a^2}{1+a}}\)

\(=\sqrt{1-a}\)

b: Khi a=24/49 thì \(P=\sqrt{1-\dfrac{24}{49}}=\sqrt{\dfrac{25}{49}}=\dfrac{5}{7}\)

c: P=2

=>1-a=4

=>a=-3

1a (đkxđ:\(x\ge0\)) \(\Leftrightarrow\dfrac{-1}{2}.\sqrt{4x}+5=0\) \(\Leftrightarrow\sqrt{4x}=10\) \(\Leftrightarrow x=25\) (t/m)

b (đkxđ:\(x\le3\) ) \(\Leftrightarrow\sqrt{3-x}\left(1-3+1,25.4\right)=6\) \(\Leftrightarrow\sqrt{3-x}=2\) \(\Leftrightarrow x=-1\) (t/m)

\(\left(1\right)=\dfrac{y}{x\left(2x-y\right)}-\dfrac{4x}{y\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(y-2x\right)\left(y+2x\right)}{xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\\ \left(2\right)=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\\ \left(3\right)=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\\ \left(4\right)=\dfrac{4x^2+15x+4+4x+7+1}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}=\dfrac{4x^2+19x+12}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}\)

\(a,=\dfrac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\dfrac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{-8x}{\left(x-2\right)^2\left(x+2\right)^2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{2x}=\dfrac{-4}{\left(x-2\right)\left(x+2\right)}\)

\(b,=\dfrac{5x^2+26xy+5y^2+5x^2-26xy+5y^2}{x\left(x-5y\right)\left(x+5y\right)}\cdot\dfrac{\left(x-5y\right)\left(x+5y\right)}{x^2+y^2}\\ =\dfrac{10\left(x^2+y^2\right)}{x\left(x^2+y^2\right)}=\dfrac{10}{x}\)

ĐKXĐ: \(x\ne\pm1\)

\(A=\left(\dfrac{\left(1+x\right)^2}{\left(1-x\right)\left(1+x\right)}-\dfrac{\left(1-x\right)^2}{\left(1-x\right)\left(1+x\right)}+\dfrac{4x^2}{\left(1-x\right)\left(1+x\right)}\right):\dfrac{4\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}\)

\(=\left(\dfrac{x^2+2x+1-\left(x^2-2x+1\right)+4x^2}{\left(1-x\right)\left(1+x\right)}\right):\dfrac{4\left(x+1\right)}{x-1}\)

\(=\left(\dfrac{4x^2+4x}{\left(1-x\right)\left(1+x\right)}\right):\dfrac{4\left(x+1\right)}{x-1}\)

\(=\dfrac{4x\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)}{4\left(x+1\right)}=-\dfrac{x}{x+1}\)

ĐKXĐ: \(x\notin\left\{-1;2;-2\right\}\)

a) Ta có: \(A=\left(\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\dfrac{2x^2+4x-1}{x^3+1}-\dfrac{1}{x+1}\right):\dfrac{x^2-4}{3x^2+6x}\)

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

\(=\left(\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{\left(x-2\right)\left(x+2\right)}{3x\left(x+2\right)}\)

\(=\dfrac{x^3+3x^2+3x+1-2x^2-4x+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}:\dfrac{x-2}{3x}\)

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

\(=\dfrac{3x}{x-2}\)

b) Để A nguyên thì \(3x⋮x-2\)

\(\Leftrightarrow3x-6+6⋮x-2\)

mà \(3x-6⋮x-2\)

nên \(6⋮x-2\)

\(\Leftrightarrow x-2\inƯ\left(6\right)\)

\(\Leftrightarrow x-2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

hay \(x\in\left\{3;1;4;0;5;-1;8;-4\right\}\)

Kết hợp ĐKXĐ, ta được:

\(x\in\left\{3;1;4;0;5;8;-4\right\}\)

Vậy: Để A nguyên thì \(x\in\left\{3;1;4;0;5;8;-4\right\}\)

Lời giải:

a) Hiển nhiên vế trái $\geq 0$ do tính chất của trị tuyệt đối.

$\Rightarrow 4x\geq 0\Rightarrow x\geq 0$. Đến đây ta có thể phá bỏ dấu trị tuyệt đối

$|x+\frac{11}{17}|+|x+\frac{2}{17}|+|x+\frac{4}{17}|=4x$

$x+\frac{11}{17}+x+\frac{2}{17}+x+\frac{4}{17}=4x$

$3x+1=4x$

$x=1$

b) Hiển nhiên vế trái $\geq 0$ nên $11x\geq 0\Rightarrow x\geq 0$

Khi đó:

$|x+\frac{1}{2}|+|x+\frac{1}{6}|+|x+\frac{1}{12}|+...+|x+\frac{1}{110}|=x+\frac{1}{2}+x+\frac{1}{6}+x+\frac{1}{12}+...+x+\frac{1}{110}$

$=10x+(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110})$

$=10x+(1-\frac{1}{11})=10x+\frac{10}{11}=11x$

$\Rightarrow x=\frac{10}{11}$

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1, bạn xem lại đề 

2, 15(x-3) + 8x-21 = 12(x+1) +120 

<=> 23x - 66 = 12x + 132 

<=> 11x = 198 <=> x = 198/11 

3, 10(3x+1) + 5 - 100 = 8(3x-1) - 6x - 4 

<=> 30x + 10 - 95 = 18x -12

<=> 12x = 73 <=> x = 73/12 

Bài 3:
a) \(\sqrt{3x-2}=4\)
⇔\(\sqrt{3x-2}=\sqrt{4^2}\)
⇔\(3x-2=4^2=16\)
    \(3x=16+2=18\)
    \(x=18:3=6\)
    Vậy \(x=6\)
b)\(\sqrt{4x^2+4x+1}-11=5\)
⇔\(\sqrt{\left(2x\right)^2+2\left(2x\right)\cdot1+1^2}-11=5\)
⇔\(\sqrt{\left(2x+1\right)^2}-11=5\)
TH1:
⇔\(\left(2x+1\right)-11=5\)
    \(2x+1=5+11=16\)
    \(2x=16-1=15\)
    \(x=15:2=7,5\)
TH2:
⇔\(\left(2x+1\right)-11=-5\)
    \(2x-1=-5+11=6\)
    \(2x=6+1=7\)
    \(x=7:2=3,5\)
    Vậy \(x=\left\{7,5;3,5\right\}\) 
    (Câu này mình không chắc chắn lắm)   
    (Học sinh lớp 6 đang làm bài này)    

Bài 4:

a: \(C=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\)

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

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

b: C-6<0

=>C<6

=>\(2\sqrt{x}< 6\)

=>\(\sqrt{x}< 3\)

=>0<=x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< x< 9\\x\ne1\end{matrix}\right.\)

Bài 3

a)\(\sqrt{3x-2}=4\Leftrightarrow3x-2=16\Leftrightarrow3x=18\Leftrightarrow x=6\)

Vậy PT có nghiệm x=6

b)\(\sqrt{4x^2+4x+1}-11=5\Leftrightarrow\sqrt{\left(2x+1\right)^2}=16\Leftrightarrow2x+1=16hoặc2x+1=-16\)

+)TH1: \(2x+1=16\Leftrightarrow x=\dfrac{15}{2}\Leftrightarrow x=7,5\)

+)TH2:\(2x+1=-16\Leftrightarrow x=\dfrac{17}{2}\Leftrightarrow x=8,5\)

Bài 4

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

\(Vậy\) \(C=2\sqrt{x}\)