Ta có:

\(P\left(0\right)=0x+b\)

\(\Rightarrow b=2007\)

\(\Rightarrow A\left(1\right)=a+b\)

Mà \(b=2007\)

\(\Rightarrow a+2007=2006\)

\(\Rightarrow a=1\)

Vậy \(P\left(x\right)=-x+2007\)

Lời giải:

$P(0)=a.0+b=b=2007$

$P(1)=a.1+b=a+b=2006$

$\Rightarrow a=2006-b=2006-2007=-1$

Vậy $a=-1; b=2007$

a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{3}{6}-\dfrac{5}{6}\)

\(\Rightarrow x+\dfrac{1}{3}=-\dfrac{2}{6}\)

\(\Rightarrow x=-\dfrac{2}{6}-\dfrac{1}{3}\)

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

b) \(\left(\dfrac{3}{8}-\dfrac{1}{5}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\)

\(\Rightarrow\left(\dfrac{15}{40}-\dfrac{8}{40}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\)

\(\Rightarrow\dfrac{7}{40}+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\)

\(\Rightarrow\dfrac{5}{8}-x=\dfrac{1}{5}-\dfrac{7}{40}\)

\(\Rightarrow\dfrac{5}{8}-x=\dfrac{1}{40}\)

\(\Rightarrow x=\dfrac{5}{8}-\dfrac{1}{40}\)

\(\Rightarrow x=\dfrac{3}{5}\)

a) x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}x+31​=21​−65​

x=-\dfrac{1}{3}-\dfrac{1}{3}x=−31​−31​

x=-\dfrac{2}{3}x=−32​;
b) \dfrac{3}{8}+\dfrac{5}{8}-\dfrac{1}{5}-x=\dfrac{1}{5}83​+85​−51​−x=51​

x=1-\dfrac{1}{5}-\dfrac{1}{5}x=1−51​−51​

x=\dfrac{3}{5}x=53​.

a) x=7-\dfrac{2}{5}+1,62=8,22x=7−52​+1,62=8,22
b) x=4 \dfrac{3}{5}+\dfrac{1}{5}-\dfrac{1}{2}=4 \dfrac{3}{10}x=453​+51​−21​=4103​
c) 2 x-x=\dfrac{3}{5}+\dfrac{4}{7}2x−x=53​+74​
x=\dfrac{41}{35}x=3541​
d) x=3 \dfrac{1}{2}-\dfrac{5}{7}+\dfrac{1}{13}-0.25x=321​−75​+131​−0.25
x=2 \dfrac{223}{364}x=2364223​

 

 x=7-\dfrac{2}{5}+1,62=8,22x=7−52​+1,62=8,22
b) x=4 \dfrac{3}{5}+\dfrac{1}{5}-\dfrac{1}{2}=4 \dfrac{3}{10}x=453​+51​−21​=4103​
c) 2 x-x=\dfrac{3}{5}+\dfrac{4}{7}2x−x=53​+74​
x=\dfrac{41}{35}x=3541​
d) x=3 \dfrac{1}{2}-\dfrac{5}{7}+\dfrac{1}{13}-0.25x=321​−75​+131​−0.25
x=2 \dfrac{223}{364}x=2364223​
 

\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)

\(\Rightarrow2x=\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{1}{10}\)

\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=2\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)

\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)

\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)

\(\Leftrightarrow x=-\dfrac{49}{8}\)

\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)

\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)

\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)

\(\Leftrightarrow x=\dfrac{413}{160}\)

a)\left(\dfrac{1}{2}+1,5\right) \cdot x=\dfrac{1}{5}(21​+1,5)⋅x=51​

2 \cdot x=\dfrac{1}{5}2⋅x=51​

x=\dfrac{1}{5}: 2x=51​:2

 x=\dfrac{1}{10} x=101​
b) \left(-1 \dfrac{3}{5}+x\right): \dfrac{12}{13}=2 \dfrac{1}{6}(−153​+x):1312​=261​

-1 \dfrac{3}{5}+x=\dfrac{13}{6} \cdot \dfrac{12}{13}−153​+x=613​⋅1312​
x=2+1 \dfrac{3}{5}x=2+153​

 x=3 \dfrac{3}{5} x=353​
c) \left(x: 2 \dfrac{1}{3}\right) \cdot \dfrac{1}{7}=\dfrac{-3}{8}(x:231​)⋅71​=8−3​

x \cdot \dfrac{3}{7} \cdot \dfrac{1}{7}=\dfrac{-3}{8}x⋅73​⋅71​=8−3​

x=\dfrac{-3}{8}: \dfrac{3}{49}x=8−3​:493​
x=\dfrac{-49}{8}=-6 \dfrac{1}{8}x=8−49​=−681​
d) \dfrac{-4}{7} \cdot x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1 \dfrac{2}{3}\right)7−4​⋅x+57​=81​:(−132​)

\dfrac{-4}{7} x+\dfrac{7}{5}=\dfrac{1}{8} \cdot \dfrac{-3}{5}7−4​x+57​=81​⋅5−3​
-\dfrac{4}{7} x=\dfrac{-3}{40}-\dfrac{7}{5} \\ x=\dfrac{-59}{40}: \dfrac{-4}{7}=\dfrac{413}{160}=2 \dfrac{93}{160}−74​x=40−3​−57​x=40−59​:7−4​=160413​=216093​
 

a) $A=\left[\frac{2}{7}\left(\frac{1}{4}-\frac{1}{3}\right)\right]:\left[\frac{2}{7}\left(\frac{1}{3}-\frac{2}{5}\right)\right]=\left(\frac{1}{4}-\frac{1}{3}\right):\left(\frac{1}{3}-\frac{2}{5}\right)=1\frac{1}{4}$.
b) $B=\frac{\frac{3}{4}\left(\frac{1}{5}-\frac{2}{7}-\frac{1}{3}+\frac{2}{7}\right)}{\frac{1}{5}\left(\frac{2}{7}+\frac{1}{3}\right)-\frac{1}{3}\left(\frac{2}{7}+\frac{1}{3}\right)}=\frac{\frac{3}{4}\left(\frac{1}{5}-\frac{1}{3}\right)}{\left(\frac{1}{5}-\frac{1}{3}\right)\left(\frac{2}{7}+\frac{1}{3}\right)}=1\frac{11}{52}$.

a) \mathrm{A}=\left[\dfrac{2}{7}\left(\dfrac{1}{4}-\dfrac{1}{3}\right)\right]:\left[\dfrac{2}{7}\left(\dfrac{1}{3}-\dfrac{2}{5}\right)\right]=\left(\dfrac{1}{4}-\dfrac{1}{3}\right):\left(\dfrac{1}{3}-\dfrac{2}{5}\right)=1 \dfrac{1}{4}A=[72​(41​−31​)]:[72​(31​−52​)]=(41​−31​):(31​−52​)=141​.
b) \mathrm{B}=\dfrac{\dfrac{3}{4}\left(\dfrac{1}{5}-\dfrac{2}{7}-\dfrac{1}{3}+\dfrac{2}{7}\right)}{\dfrac{1}{5}\left(\dfrac{2}{7}+\dfrac{1}{3}\right)-\dfrac{1}{3}\left(\dfrac{2}{7}+\dfrac{1}{3}\right)}=\dfrac{\dfrac{3}{4}\left(\dfrac{1}{5}-\dfrac{1}{3}\right)}{\left(\dfrac{1}{5}-\dfrac{1}{3}\right)\left(\dfrac{2}{7}+\dfrac{1}{3}\right)}=1 \dfrac{11}{52}B=51​(72​+31​)−31​(72​+31​)43​(51​−72​−31​+72​)​=(51​−31​)(72​+31​)43​(51​−31​)​=15211​

a) $A=\frac{3}{5}.\frac{6}{7}+\frac{3}{7}.\frac{3}{5}-\frac{2}{7}.\frac{3}{5}$
$=\frac{3}{5}\cdot \left(\frac{6}{7}+\frac{3}{7}-\frac{2}{7}\right)=\frac{3}{5}$
b) $B=\left(-13\cdot \frac{2}{5}+\frac{-2}{9}\cdot \frac{2}{5}+\frac{2}{5}\cdot \frac{11}{9}\right)\cdot \frac{5}{2}$
$=\left(-13-\frac{2}{9}+\frac{11}{9}\right)\cdot \frac{2}{5}\cdot \frac{5}{2}=-13+\left(\frac{11}{9}-\frac{2}{9}\right)=-12.$
c) $C=\left(\frac{-4}{5}+\frac{5}{7}\right)\cdot \frac{3}{2}+\left(\frac{-1}{5}+\frac{2}{7}\right)\cdot \frac{3}{2}=\left(\frac{-4}{5}+\frac{5}{7}+\frac{-1}{5}+\frac{2}{7}\right)\cdot \frac{3}{2}=\left(\left(\frac{-4}{5}+\frac{-1}{5}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\right)\cdot \frac{3}{2}=0.$
d) $D=\frac{4}{9}:\left(\frac{1}{15}-\frac{10}{15}\right)+\frac{4}{9}:\left(\frac{2}{22}-\frac{5}{22}\right)$
$=\frac{4}{9}:\frac{-3}{5}+\frac{4}{9}:\frac{-3}{22}=\frac{4}{9}\cdot \frac{-5}{3}+\frac{4}{9}.\frac{-22}{3}$
$=\frac{4}{9}\cdot \left(\frac{-5}{3}+\frac{-22}{3}\right)=\frac{4}{9}.\frac{-27}{3}=-4.$

a) \mathrm{A}=\dfrac{3}{5}. \dfrac{6}{7}+\dfrac{3}{7}. \dfrac{3}{5}-\dfrac{2}{7}. \dfrac{3}{5}A=53​. 76​+73​. 53​−72​. 53​

b)  \mathrm{B} =\left(-13 \cdot \dfrac{2}{5}+\dfrac{-2}{9} \cdot \dfrac{2}{5}+\dfrac{2}{5} \cdot \dfrac{11}{9}\right) \cdot \dfrac{5}{2} B=(−13⋅52​+9−2​⋅52​+52​⋅911​)⋅25​
=\left(-13-\dfrac{2}{9}+\dfrac{11}{9}\right) \cdot \dfrac{2}{5} \cdot \dfrac{5}{2}=-13+\left(\dfrac{11}{9}-\dfrac{2}{9}\right)=-12 .=(−13−92​+911​)⋅52​⋅25​=−13+(911​−92​)=−12.
c) \mathrm{C} =\left(\dfrac{-4}{5}+\dfrac{5}{7}\right) \cdot \dfrac{3}{2}+\left(\dfrac{-1}{5}+\dfrac{2}{7}\right) \cdot \dfrac{3}{2} =\left(\dfrac{-4}{5}+\dfrac{5}{7}+\dfrac{-1}{5}+\dfrac{2}{7}\right) \cdot \dfrac{3}{2}=\left(\left(\dfrac{-4}{5}+\dfrac{-1}{5}\right)+\left(\dfrac{5}{7}+\dfrac{2}{7}\right)\right) \cdot \dfrac{3}{2}=0 .C=(5−4​+75​)⋅23​+(5−1​+72​)⋅23​=(5−4​+75​+5−1​+72​)⋅23​=((5−4​+5−1​)+(75​+72​))⋅23​=0.
d) \mathrm{D}=\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{10}{15}\right)+\dfrac{4}{9}:\left(\dfrac{2}{22}-\dfrac{5}{22}\right)D=94​:(151​−1510​)+94​:(222​−225​)

a) $P=\frac{2}{3}+\frac{1}{4}+\frac{3}{5}-\frac{7}{45}+\frac{5}{9}+\frac{1}{12}+\frac{1}{35}$ $=\left(\frac{2}{3}+\frac{1}{4}+\frac{1}{12}\right)+\left(\frac{5}{9}-\frac{7}{45}\right)+\frac{3}{5}+\frac{1}{35}=1+\frac{4}{5}+\frac{3}{5}+\frac{1}{35}=2\frac{1}{35}$

$\mathrm{a)\; P}=\frac{2}{3}+\frac{1}{4}+\frac{3}{5}-\frac{7}{45}+\frac{5}{9}+\frac{1}{12}+\frac{1}{35}$ 

    P $=\left(\frac{2}{3}+\frac{1}{4}+\frac{1}{12}\right)+\left(\frac{5}{9}-\frac{7}{45}\right)+\frac{3}{5}+\frac{1}{35}$

P =$1+\frac{4}{5}+\frac{3}{5}+\frac{1}{35}=2\frac{1}{35}$.

b)

$Q\; =\left(5-\frac{3}{4}+\frac{1}{5}\right)-\left(6+\frac{7}{4}-\frac{8}{5}\right)-\left(2-\frac{5}{4}+\frac{16}{5}\right)$

$Q=(5-6-2)+\left(-\frac{3}{4}-\frac{7}{4}+\frac{5}{4}\right)+(\frac{1}{5}-\frac{8}{5}-\frac{16}{5}$)

$Q\; =\; -\left(3+\frac{5}{4}+\frac{23}{5}\right)$ $=-(3+1\frac{1}{4}+4\frac{3}{5}$

$Q\; =-8\frac{17}{20}$.

a) $A=\left(\frac{1}{3}+\frac{2}{3}\right)-\left(\frac{8}{15}+\frac{7}{15}\right)+\left(\frac{-1}{7}+1\frac{1}{7}\right)=1-1+1=1$;
b) $B=\left(0.25-1\frac{1}{4}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)-\frac{1}{8}$
$=\left(\frac{1}{4}-1-\frac{1}{4}\right)+1-\frac{1}{8}=\frac{-1}{8}$.