a: \(\left|x+\dfrac{4}{15}\right|-\left|-3.75\right|=-\left|-2.5\right|\)

\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.5+3.75=1.25=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=\dfrac{5}{4}\\x+\dfrac{4}{15}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{59}{60}\\x=-\dfrac{91}{60}\end{matrix}\right.\)

c: \(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)

d: Ta có: \(\left|x\left(x^2-\dfrac{5}{4}\right)\right|=x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x\left(x^2-\dfrac{5}{4}\right)=x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x\left(x^2-\dfrac{9}{4}\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;\dfrac{3}{2}\right\}\)

Ta có \(A\left(1\right)=a+b+c+d=A\left(-1\right)=a+b-c+d\)

=> c = -c (1)

Ta có \(A\left(2\right)=16+4b+2c+d=A\left(-2\right)=16a+4b-2c+d\)

=> 2c = -2c (2)

Từ (1) và (2) => A(x) = A(-x)

B1:a )

(x-4).(y+3)=-3=-1.3=-3.1

ta có bảng sau:

a, => x= -9. 2

=> x= -18

b, => x= 12. (-4)

=> x=-48

c, => 152-(3x-1)=54

=> (3x-1)= 152- 54= 98

=> 3x= 98+1= 99

=> x = 99:3

=> x = 33

tick cho mik nha

Ta có:

a) \(x:2=-9\Rightarrow x=\left(-9\right).2=-18\)

b)\(x:\left(-4\right)=12\Rightarrow x=12.\left(-4\right)=-48\)

c)\(x+x+80=-8\Rightarrow2x+80=-8\Rightarrow2x=-8-80=-88\Rightarrow x=\left(-88\right):2=-44\)d)\(152-\left(3x-1\right)=\left(-2\right).\left(-27\right)=54\Rightarrow152-54=3x-1\Rightarrow98=3x-1\Rightarrow3x=99\Rightarrow x=33\)

a)

\(\left(\frac{x-1}{4}\right)^2=\frac{4}{9}\)

⇒ \(\frac{x-1}{4}=\frac{2}{3}\)

\(\Rightarrow\frac{\left(x-1\right).3}{12}=\frac{8}{12}\)

=> (x - 1) = 8/3

=> x = 8/3 - 1

=> x = 5/3

d) x : -9/2 = -4/25

x = -4/25 . -9/2

x = 18/25

c) (x - 3/5)^2 = 4

=> x - 3/5 = 2

=> x = 2 + 3/5

=> x = 13/5

a,\(\dfrac{1}{9}\)=\(\dfrac{x}{27}\)

\(\dfrac{3}{27}\)= \(\dfrac{x}{27}\)

3=x

Vậy x =3

b,\(\dfrac{4}{x}\)=\(\dfrac{8}{6}\)

\(\dfrac{4}{x}\)=\(\dfrac{4}{3}\)

x=3

Vậy x = 3

d,\(\dfrac{x}{-2}\)=\(\dfrac{-8}{x}\)

x.x = (-2).(-8)

x² =16

x² = 4²

x = 4

Vậy x = 4

Bài 1:

A.\(\left(\sqrt{x}+2\right)\) = -1 (ĐK: \(x\ge0\)

\(\Leftrightarrow\dfrac{1}{x-4}\left(\sqrt{x}+2\right)=-1\)

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

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

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

\(\Leftrightarrow\sqrt{x}=1\\ \Leftrightarrow x=1\left(TM\right)\)

Vậy x = 1

Bài 2: ĐK: \(x\ge0\)

Để \(B\in Z\Leftrightarrow\dfrac{3}{\sqrt{x}+2}\in Z\Leftrightarrow\sqrt{x}+2\inƯ\left(3\right)\)\(\Leftrightarrow\sqrt{x}+2\in\left\{\pm1,\pm3\right\}\)\(\Leftrightarrow x\in\left\{1\right\}\)

Bài 3:

a, Ta có: \(x+\sqrt{x}+1=x+2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}+1\\ =\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

Ta có: 2 > 0 và \(x+\sqrt{x}+1>0\Rightarrow C>0\) và \(x\ne1\)

b, ĐK: \(x\ge0,x\ne1\)

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

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

Ta có: \(\sqrt{x}\ge0\forall x\Rightarrow\sqrt{x}+\dfrac{1}{2}\ge\dfrac{1}{2}\forall x\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2\ge\dfrac{1}{4}\)

\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge1\Leftrightarrow\dfrac{2}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le2\)

Dấu bằng xảy ra \(\Leftrightarrow\sqrt{x}+\dfrac{1}{2}=\dfrac{1}{2}\\ \Leftrightarrow x=0\left(TM\right)\)

Vậy MaxC = 2 khi x = 0

Còn cái GTNN chưa tính ra được, để sau nha

Bài 4: ĐK: \(x\ge0,x\ne1\)

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

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

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

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

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

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

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

\(D=3\Leftrightarrow\sqrt{x}-1=3\Leftrightarrow x=2\left(TM\right)\)

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

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

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

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

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3-\sqrt{x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(L\right)\\x=9\left(TM\right)\end{matrix}\right.\)

Bài 5: \(E< -1\Leftrightarrow\dfrac{-3x}{2x+4\sqrt{x}}< -1\)\(\Leftrightarrow\dfrac{-3x}{2x+4\sqrt{x}}+1< 0\Leftrightarrow\dfrac{-3x+2x+4\sqrt{x}}{2x+4\sqrt{x}}< 0\)

\(\Leftrightarrow\dfrac{4\sqrt{x}-x}{2x+4\sqrt{x}}< 0\Leftrightarrow\dfrac{\sqrt{x}\left(4-\sqrt{x}\right)}{2x+4\sqrt{x}}< 0\)

Ta có: \(\sqrt{x}>0\Leftrightarrow x>0\Leftrightarrow2x+4\sqrt{x}>0\) mà \(\dfrac{\sqrt{x}\left(4-\sqrt{x}\right)}{2x+4\sqrt{x}}< 0\)\(\Rightarrow\sqrt{x}\left(4-\sqrt{x}\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}< 0\left(L\right)\\4-\sqrt{x}>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}>0\\4-\sqrt{x}< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x< 16,x\ne0\\\left\{{}\begin{matrix}x>0\\x< 16\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< 16,x\ne0\\0< x< 16\end{matrix}\right.\)