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=[27(14−13)]:[27(13−25)]=(14−13):(13−25)=114A=[27(14−13)]:[27(13−25)]=(14−13):(13−25)=114.
b) B=34(15−27−13+27)15(27+13)−13(27+13)=34(15−13)(15−13)(27+13)=11152B=34(15−27−13+27)15(27+13)−13(27+13)=34(15−13)(15−13)(27+13)=11152.

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=35.67+37.35−27.35A=35.67+37.35−27.35
=35⋅(67+37−27)=35=35⋅(67+37−27)=35
b) B=(−13⋅25+−29⋅25+25⋅119)⋅52B=(−13⋅25+−29⋅25+25⋅119)⋅52
=(−13−29+119)⋅25⋅52=−13+(119−29)=−12.=(−13−29+119)⋅25⋅52=−13+(119−29)=−12.
c) C=(−45+57)⋅32+(−15+27)⋅32=(−45+57+−15+27)⋅32=((−45+−15)+(57+27))⋅32=0.C=(−45+57)⋅32+(−15+27)⋅32=(−45+57+−15+27)⋅32=((−45+−15)+(57+27))⋅32=0.
d) D=49:(115−1015)+49:(222−522)D=49:(115−1015)+49:(222−522)
=49:−35+49:−322=49⋅−53+49.−223=49:−35+49:−322=49⋅−53+49.−223
=49⋅(−53+−223)=49.−273=−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=23+14+35−745+59+112+135P=23+14+35−745+59+112+135 =(23+14+112)+(59−745)+35+135=1+45+35+135=2135

a) P=23+14+35−745+59+112+135P=23+14+35−745+59+112+135 

    P =(23+14+112)+(59−745)+35+135

P =1+45+35+135=2135=(23+14+112)+(59−745)+35+135=1+45+35+135=2135.

b)

Q =(5−34+15)−(6+74−85)−(2−54+165)

Q=(5−6−2)+(−34−74+54)+(15−85−165Q=(5−34+15)−(6+74−85)−(2−54+165))

Q = −(3+54+235)Q=(5−6−2)+(−34−74+54)+(15−85−165)=−(3+54+235) =−(3+114+435

Q =−81720=−(3+114+435)=−81720.

a) A=(13+23)−(815+715)+(−17+117)=1−1+1=1A=(13+23)−(815+715)+(−17+117)=1−1+1=1;
b) B=(0.25−114)+(35+25)−18B=(0.25−114)+(35+25)−18
=(14−1−14)+1−18=−18=(14−1−14)+1−18=−18.

\(\Leftrightarrow40+2xy=x\left(x\ne0\right)\)

\(\Leftrightarrow x\left(1-2y\right)=40\Leftrightarrow x=\dfrac{40}{1-2y}\)

Do 2y chẵn => 1-2y lẻ

Để x nguyên thì 1-2y là ước của 40

\(\Rightarrow1-2y=\left\{-5;-1;1;5\right\}\Rightarrow y=\left\{3;1;0;-2\right\}\)

\(\Rightarrow x=\left\{-8;-40;40;8\right\}\)

M N P B A H I

a/

Xét tg MAH và tg BAN có

AM=AB (gt); AN=AH (gt)

\(\widehat{MAH}=\widehat{BAN}\) (góc đối đỉnh)

=> tg MAH = tg BAN (c.g.c)

b/

Ta có tg MAH = tg BAN (cmt) mà \(\Rightarrow\widehat{BNA=}\widehat{MHA}=90^o\)

Xét tg vuông BAN có AB>BN (trong tg vuông cạnh huyền là cạnh có số đo lớn nhất)

Mà AB=AM

=> AM>BN (1)

Xét tg vuông MAH có \(\widehat{MAH}\) là góc nhọn => \(\widehat{MAN}\) là góc tù

Xét tg MAN có MN>AM (trong tg cạnh đối diện với góc tù là cạnh có số đo lớn nhất) (2)

Từ (1) và (2) => MN>BN

Ta có tg MAH = tg BAN (cmt) => \(\widehat{NBM}=\widehat{AMH}\) (3)

Xét tg BMN có

MN>BN (cmt) => \(\widehat{NBM}>\widehat{NMA}\) (trong tg góc đối diện với cạnh có số đo lớn hơn thì lớn hơn góc đối diện với cạnh có số đo nhỏ hơn) (4)

Từ (3) và (4) => \(\widehat{AMH}>\widehat{NMA}\)

c/

Ta có \(\widehat{BNA}=90^o\left(cmt\right)\Rightarrow BN\perp NP\) (1)

Xét tg MNP có \(MH\perp NP\left(gt\right)\) => MH là đường cao

=> MH là đường trung tuyến của tg MNP (trong tg cân đường cao hạ từ đỉnh đồng thời là đường trung tuyến) => HN=HP

Mà IB=IP (gt)

=> IH là đường trung bình của tg BNP => IH//BN (2)

Từ (1) và (2) => \(IH\perp NP\) mà \(MH\perp NP\)

=> M; H; I thảng hàng (từ 1 điểm trên đường thẳng chỉ dựng được duy nhất 1 đường thẳng vuông góc với đường thẳng đã cho)

Xét tg INP có

\(IH\perp NP\) => IH là đường cao của tg INP

HN=HP (cmt) => IH là đường trung tuyến của tg INP

=> tg INP là tg cân tại I (trong tg đường cao đồng thời là đường trung tuyến thì tg đó là tg cân) => IN=IP (cạn bên tg cân)

Mà IP=IB (gt) và IP+IB=BP

=> IN=1/2BP