(2x - 1)/3 -(x +2)/6 =5

2(2x - 1)/6 -  (x+2)/6 = (5 .6)/6

4x -2 -x -2 = 30

3x = 34

x = 34/3

Đang giải phương trình mà bn

Đường thẳng y=x-4 có a=1>0

nên đường thẳng y=x-4 tạo với trục Ox một góc nhọn có số đo là:

\(arctan\left(1\right)=45^0\)

a: Xét ΔABC vuông tại A và ΔHAC vuông tại H có

\(\widehat{ACB}\) chung

Do đó: ΔABC~ΔHAC

b: ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(AC=\sqrt{10^2-8^2}=6\left(cm\right)\)

ΔABC~ΔHAC
=>\(\dfrac{CA}{CH}=\dfrac{CB}{CA}\)

=>\(CA^2=CH\cdot CB\)

=>\(CH\cdot10=36\)

=>CH=36/10=3,6(cm)

ta có: HB+HC=BC

=>HB+3,6=10

=>HB=6,4(cm)

\(\dfrac{x+1}{2024}+\dfrac{x+2}{2023}=\dfrac{x+3}{2022}+\dfrac{x+4}{2021}\)

=>\(\left(\dfrac{x+1}{2024}+1\right)+\left(\dfrac{x+2}{2023}+1\right)=\left(\dfrac{x+3}{2022}+1\right)+\left(\dfrac{x+4}{2021}+1\right)\)

=>\(\dfrac{x+2025}{2024}+\dfrac{x+2025}{2023}=\dfrac{x+2025}{2022}+\dfrac{x+2025}{2021}\)

=>\(\left(x+2025\right)\left(\dfrac{1}{2024}+\dfrac{1}{2023}-\dfrac{1}{2022}-\dfrac{1}{2021}\right)=0\)

=>x+2025=0

=>x=-2025

a: Xét ΔABC vuông tại A và ΔHAC vuông tại H có

\(\widehat{ACB}\) chung

Do đó: ΔABC~ΔHAC

Xét ΔHAB vuông tại H và ΔHCA vuông tại H có

\(\widehat{HAB}=\widehat{HCA}\left(=90^0-\widehat{ABC}\right)\)

Do đó: ΔHAB~ΔHCA

=>\(\dfrac{HA}{HC}=\dfrac{HB}{HA}\)

=>\(HA^2=HB\cdot HC\)

b: Xét ΔEAB vuông tại A và ΔEFC vuông tại F có

\(\widehat{AEB}=\widehat{FEC}\)(hai góc đối đỉnh)

Do đó: ΔEAB~ΔEFC

=>\(\dfrac{EA}{EF}=\dfrac{EB}{EC}\)

=>\(EA\cdot EC=EB\cdot EF\)

a: Xét ΔHBA vuông tại H và ΔABC vuông tại A có

\(\widehat{HBA}\) chung

Do đó: ΔHBA~ΔABC

=>\(\dfrac{AB}{BC}=\dfrac{HA}{AC}\)

=>\(AB\cdot AC=HA\cdot BC\)

b: Xét ΔAMH vuông tại M và ΔAHB vuông tại H có

\(\widehat{MAH}\) chung

Do đó: ΔAMH~ΔAHB

=>\(\dfrac{AM}{AH}=\dfrac{AH}{AB}\)

=>\(AH^2=AM\cdot AB\left(1\right)\)

Xét ΔANH vuông tại N và ΔAHC vuông tại H có

\(\widehat{NAH}\) chung

Do đó: ΔANH~ΔAHC

=>\(\dfrac{AN}{AH}=\dfrac{AH}{AC}\)

=>\(AN\cdot AC=AH^2\left(2\right)\)

Từ (1),(2) suy ra \(AM\cdot AB=AN\cdot AC\)

=>\(\dfrac{AM}{AC}=\dfrac{AN}{AB}\)

Xét ΔAMN vuông tạiA và ΔACB vuông tại A có

\(\dfrac{AM}{AC}=\dfrac{AN}{AB}\)

Do đó: ΔAMN~ΔACB

c:

ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(BC=\sqrt{6^2+8^2}=10\left(cm\right)\)

\(AH\cdot BC=AB\cdot AC\)

=>\(AH\cdot10=6\cdot8=48\)

=>AH=48/10=4,8(cm)

ΔABC vuông tại A

=>\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC=\dfrac{1}{2}\cdot6\cdot8=24\left(cm^2\right)\)

Xét tứ giác AMHN có \(\widehat{AMH}=\widehat{ANH}=\widehat{MAN}=90^0\)

nên AMHN là hình chữ nhật

=>MN=AH

mà AH=4,8cm

nên MN=4,8(cm)

ΔAMN~ΔACB

=>\(\dfrac{S_{AMN}}{S_{ACB}}=\left(\dfrac{MN}{CB}\right)^2=\left(\dfrac{4.8}{10}\right)^2\)

=>\(S_{AMN}=\left(\dfrac{4.8}{10}\right)^2\cdot24=5,5296\left(cm^2\right)\)

\(S_{AMN}+S_{BMNC}=S_{ABC}\)

=>\(S_{BMNC}=24-5,5296=18,4704\left(cm^2\right)\)

`#3107.101107`

`a)`

`5 - (x - 6) = 4(2x - 3)`

\(\Leftrightarrow5-x+6=8x-12\)

\(\Leftrightarrow-x-8x=-12-5-6\)

\(\Leftrightarrow-9x=-23\)

\(\Leftrightarrow x=\dfrac{-23}{-9}=\dfrac{23}{9}\)

Vậy, \(x=\dfrac{23}{9}\) 

`b)`

\(5\left(3x+2\right)-4\left(5-3x\right)=1\)

\(\Leftrightarrow15x+10-20+12x=1\)

\(\Leftrightarrow27x-10=1\)

\(\Leftrightarrow27x=11\)

\(\Leftrightarrow x=\dfrac{11}{27}\)

Vậy, \(x=\dfrac{11}{27}\)

`c)`

\(-4\left(x-3\right)=6x+\left(x-3\right)\)

\(\Leftrightarrow-4x+12=6x+x-3\)

\(\Leftrightarrow-4x-7x=-12-3\)

\(\Leftrightarrow11x=-15\)

\(\Leftrightarrow x=\dfrac{-15}{11}\)

Vậy, \(x=-\dfrac{15}{11}\)

`d)`

\(\dfrac{x}{3}-\dfrac{5x}{6}-\dfrac{5x}{12}=\dfrac{x}{4}-5\)

\(\Leftrightarrow\dfrac{4x}{12}-\dfrac{10x}{12}-\dfrac{5x}{12}=\dfrac{3x}{12}-\dfrac{60}{12}\)

\(\Leftrightarrow4x-10x-5x=3x-60\)

\(\Leftrightarrow-11x-3x=-60\)

\(\Leftrightarrow-14x=-60\)

\(\Leftrightarrow x=\dfrac{-60}{-14}=\dfrac{30}{7}\)

Vậy, \(x=\dfrac{30}{7}\) 

`e)`

\(\dfrac{x-1}{2}-\dfrac{x+1}{15}-\dfrac{2x-13}{6}=0\)

\(\Leftrightarrow\dfrac{15\left(x-1\right)}{30}-\dfrac{2\left(x+1\right)}{30}-\dfrac{5\left(2x-13\right)}{30}=0\)

\(\Leftrightarrow15\left(x-1\right)-2\left(x+1\right)-5\left(2x-13\right)=0\)

\(\Leftrightarrow15x-15-2x-2-10x+65=0\)

\(\Leftrightarrow3x+48=0\)

\(\Leftrightarrow3x=-48\)

\(\Leftrightarrow x=-16\)

Vậy, \(x=-16.\)

`f)`

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

\(\Leftrightarrow\dfrac{9\left(3-x\right)}{24}+\dfrac{16\left(5-x\right)}{24}=\dfrac{12\left(1-x\right)}{24}-\dfrac{48}{24}\)

\(\Leftrightarrow27-9x+80-16x=12-12x-48\)

\(\Leftrightarrow-9x-16x+12x=12-48-27-80\)

\(\Leftrightarrow-13x=-143\)

\(\Leftrightarrow x=11\)

Vậy, `x = 11.`

`g)`

\(\dfrac{3\left(5x-2\right)}{4}-2=\dfrac{7x}{3}-5\left(x-7\right)\)

\(\Leftrightarrow\dfrac{9\left(5x-2\right)}{12}-\dfrac{24}{12}=\dfrac{28x}{12}-\dfrac{60\left(x-7\right)}{12}\)

\(\Leftrightarrow45x-18-24=28x-60x+420\)

\(\Leftrightarrow45x-28x+60x=42+420\)

\(\Leftrightarrow77x=462\)

\(\Leftrightarrow x=6\)

Vậy, `x = 6`

`h)`

\(\dfrac{x+5}{2}+\dfrac{3-2x}{4}=x-\dfrac{7+x}{6}\)

\(\Leftrightarrow\dfrac{6\left(x+5\right)}{12}+\dfrac{3\left(3-2x\right)}{12}=\dfrac{12x}{12}-\dfrac{2\left(7+x\right)}{12}\)

\(\Leftrightarrow6x+30+9-6x=12x-14-2x\)

\(\Leftrightarrow-10x=-53\)

\(\Leftrightarrow x=5,3\)

Vậy, `x = 5,3`

`i)`

\(\dfrac{x-3}{11}+\dfrac{x+1}{3}=\dfrac{x+7}{9}-1\)

\(\Leftrightarrow\dfrac{9\left(x-3\right)}{99}+\dfrac{33\left(x+1\right)}{99}=\dfrac{11\left(x+7\right)}{99}-\dfrac{99}{99}\)

\(\Leftrightarrow9x-27+33x+33=11x+77-99\)

\(\Leftrightarrow42x-11x=-22-6\)

\(\Leftrightarrow31x=-28\)

\(\Leftrightarrow x=-\dfrac{28}{31}\)

Vậy, \(x=-\dfrac{28}{31}.\)

a: 5+2x=20-3x

=>2x+3x=20-5

=>3x=15

=>\(x=\dfrac{15}{3}=5\)

b: \(7-\left(2x+4\right)=-\left(x+4\right)\)

=>\(7-2x-4+x+4=0\)

=>7-x=0

=>x=7

c: \(\left(5x+2\right)-3\left(2x+1\right)=-3x+7\)

=>\(5x+2-6x-3=-3x+7\)

=>-3x+7=-x-1

=>-2x=-8

=>x=4

d: \(2x-\left(3-5x\right)=4\left(x+3\right)\)

=>2x-3+5x=4x+12

=>7x-3=4x+12

=>3x=15

=>\(x=\dfrac{15}{3}=5\)

e: \(5\left(2x-3\right)-4\left(5x-7\right)=19-2\left(x+11\right)\)

=>\(10x-15-20x+28=19-2x-22\)

=>\(-10x+13=-2x-3\)

=>-8x=-16

=>x=2

f: \(x^2-\left(x+2\right)\left(x-2\right)=2x\)

=>\(x^2-\left(x^2-4\right)=2x\)

=>\(2x=x^2-x^2+4=4\)

=>\(x=\dfrac{4}{2}=2\)

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

=>\(\dfrac{1-3x+6\left(x-1\right)}{6}=\dfrac{3x+6}{6}\)

=>1-3x+6x-6=3x+6

=>3x-5=3x+6

=>-5=6(loại)

h: \(\dfrac{3\left(2x+1\right)}{4}-5-\dfrac{3x+2}{10}=\dfrac{2\left(3x-1\right)}{5}\)

=>\(\dfrac{15\left(2x+1\right)-100-2\left(3x+2\right)}{20}=\dfrac{8\left(3x-1\right)}{20}\)

=>30x+15-100-6x-4=24x-8

=>24x-89=24x-8

=>-89=-8(vô lý)

a: 2x+6=0

=>2x=-6

=>x=-3

b: 4x+20=0

=>4x=-20

=>x=-5

c: 2(x+1)=5x-7

=>5x-7=2x+2

=>3x=9

=>x=3

d: \(2x-3=0\)

=>2x=3

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

e: 3x-1=x+3

=>3x-x=3+1

=>2x=4

=>\(x=\dfrac{4}{2}=2\)

f: 15-7x=9-3x

=>-7x+3x=9-15

=>-4x=-6

=>x=1,5

g: x-3=18

=>x=18+3=21

h: 2x+1=15-5x

=>2x+5x=15-1

=>7x=14

=>\(x=\dfrac{14}{7}=2\)

i: 3x-2=2x+5

=>3x-2x=2+5

=>x=7

k: -4x+8=0

=>-4x=-8

=>\(x=\dfrac{-8}{-4}=2\)

l: 2x+3=0

=>2x=-3

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

m: 4x+5=3x

=>4x-3x=-5

=>x=-5

a: |x-4|=1

=>\(\left[{}\begin{matrix}x-4=1\\x-4=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)

Khi x=5 thì \(B=\dfrac{5+1}{5-3}=\dfrac{6}{2}=3\)

b: \(A=\dfrac{x}{x-3}-\dfrac{x+1}{x+3}+\dfrac{3x-3}{9-x^2}\)

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

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

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

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

c: \(M=B:A=\dfrac{2}{x-3}:\dfrac{x}{x-3}=\dfrac{2}{x}\)

Để M=5 thì 2/x=5

=>\(x=\dfrac{2}{5}\left(nhận\right)\)

d: \(N=B-A=\dfrac{2}{x-3}-\dfrac{x}{x-3}=\dfrac{2-x}{x-3}\)

Để N là số nguyên thì \(2-x⋮x-3\)

=>\(x-2⋮x-3\)

=>\(x-3+1⋮x-3\)

=>\(1⋮x-3\)

=>\(x-3\in\left\{1;-1\right\}\)

=>\(x\in\left\{4;2\right\}\)