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\(\frac{5}{3}-\frac{1}{4}+\frac{1}{3}-\frac{3}{4}\)

\(=\frac{5}{3}+\frac{-1}{4}+\frac{1}{3}+\frac{-3}{4}\)

\(=\left(\frac{5}{3}+\frac{1}{3}\right)+\left(\frac{-1}{4}+\frac{-3}{4}\right)\)

\(=\frac{6}{3}+\frac{-4}{4}\)

\(=2+\left(-1\right)\)

\(=1\)

a: \(A=\dfrac{2x+2}{5}\)

b: \(A=\dfrac{x+1+x}{x\left(x+1\right)}=\dfrac{2x+1}{x\left(x+1\right)}\)

c: \(A=\dfrac{4x+5+9x-5}{2x-1}=\dfrac{14x}{2x-1}\)

a: \(A=\dfrac{2x+5+x-2}{3}=\dfrac{3x+3}{3}=x+1\)

b: \(A=\dfrac{x-1}{x-1}=1\)

1a)  x + 2/5 = 1/2 =5/10-4/10=1/10

b=2/7+2/5=10/35+14/35=24/35

c)Xx3/4=9/10

         x=9/10:3/4

        x36/30=6/5

\(Ba\rightarrow Ba^{2+}+2e\\ K\rightarrow K^++e\\ O+2e\rightarrow O^{2-}\\ N\rightarrow N^{5+}+5e\)

a) \(P=\dfrac{\sqrt{x}+5}{\sqrt{x}-2}=\dfrac{\sqrt{9}+5}{\sqrt{9}-2}=\dfrac{3+5}{3-2}=8\)

b) \(Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{5\sqrt{x}-2}{4-x}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

c) \(M=\dfrac{Q}{P}=\dfrac{\sqrt{x}}{\sqrt{x}-2}:\dfrac{\sqrt{x}+5}{\sqrt{x}-2}=\dfrac{\sqrt{x}}{\sqrt{x}-2}.\dfrac{\sqrt{x}-2}{\sqrt{x}+5}=\dfrac{\sqrt{x}}{\sqrt{x}+5}< \dfrac{1}{2}\)

\(\Leftrightarrow2\sqrt{x}< 3\sqrt{x}+15\Leftrightarrow\sqrt{x}>-15\left(đúng\forall x\ge0,x\ne4\right)\)

d) \(M=\dfrac{\sqrt{x}}{\sqrt{x}+5}=1-\dfrac{5}{\sqrt{x}+5}\in Z\)

\(\Rightarrow\sqrt{x}+5\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)

Do \(x\ge0,x\ne4\)

\(\Rightarrow x\in\left\{0\right\}\)

a: Xét ΔIMC vuông tại I và ΔINC vuông tại I có 

IM=IN

CI chung

Do đó: ΔIMC=ΔINC

b: Xét ΔCKB có 

M là trung điểm của BC

MN//KB

Do đó: N là trung điểm của CK

a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)

\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{-5}{12}\)

b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{5}\)

c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)

\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-77}{120}\)

d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)

\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{-7}{20}\)

e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-59}{105}\)

g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-13}{12}\)

a, 5x(x-2) + (2-x)=0
⇔5x(x-2) - (x-2) =0
⇔(x-2)(5x-1)=0
\(\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy....
c, (x³ - x²) - 4x² + 8x -4 =0
⇔x³ - x² -4x² + 8x - 4=0
⇔x²(x-1) - 4x(x-1) +4(x-1) =0
⇔(x-1) (x-2)²=0
⇔\(\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy...
Phần b cậu có chép sai đề không?