16:Sửa đề: \(\frac12x+\frac16\left(x-2\right)=\frac34-2x\)

=>\(\frac12x+\frac16x-\frac26=\frac34-2x\)

=>\(\frac46x-\frac13=\frac34-2x\)

=>\(\frac23x+2x=\frac34+\frac13\)

=>\(\frac83x=\frac{9}{12}+\frac{4}{12}=\frac{13}{12}\)

=>\(x=\frac{13}{12}:\frac83=\frac{13}{12}\times\frac38=\frac{13}{4\times8}=\frac{13}{32}\)

19: \(\frac{5}{12}x+3=\frac13-\frac{7}{12}x\)

=>\(\frac{5}{12}x+\frac{7}{12}x=\frac13-3\)

=>\(\frac{12}{12}x=\frac13-\frac93=-\frac83\)

=>\(x=-\frac83\)

20: \(\frac12x+\frac52=\frac72x-\frac34\)

=>\(\frac12x-\frac72x=-\frac34-\frac52\)

=>\(-3x=-\frac34-\frac{10}{4}=-\frac{13}{4}\)

=>\(3x=\frac{13}{4}\)

=>\(x=\frac{13}{4}:3=\frac{13}{12}\)

\(\dfrac{3}{4}+\dfrac{1}{4}.x=-\dfrac{1}{2}\)

\(\dfrac{1}{4}.x=-\dfrac{1}{2}-\dfrac{3}{4}\)

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

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

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

\(x=-\dfrac{20}{4}=-5\)

1)

\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)

Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:

\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)

2) Bạn xem lại đề!

d

Bài 2:

a)|x| < 3

x\(\in\){-2;-1;0;1;2}

b)|x - 4 | < 3

x\(\in\){ 6 ; 5 ; 4 ; 3 ; 2 }

c) | x + 10 | < 2

x\(\in\){ -2 ; -10 }

Bài 1:

A = 1 + 2 - 3 + 4 + 5 - 6 +...+98 - 99

A = (1 + 4 + 7 +...+97) + [(2-3)+(5-6)+...+(98-99)]

A = 1617 + [(-1)+(-1)+...+(-1)]

A = 1617 + (-49)

A = +(1617-49) = A = 1568

B = - 2 - 4 + 6 - 8 + 10 + 12 - .... + 60

B =  

2) 

a) \(x\in\left\{2;1;0;-1;-2\right\}\)

b) \(x\in\left\{6;-6;5;-5;4\right\}\)

c) \(x\in\left\{-9;-11;-10\right\}\)

3)

\(\left(a;b\right)\in\left\{\left(0;1\right);\left(0;-1\right);\left(1;0\right);\left(-1;0\right)\right\}\)

Emma Granger: Cảm ơn bạn nhìu ạ ! :))

1/2x + 1/5 = 2/3x - 1/4

=> 1/2x - 2/3x = -1/4 - 1/5

=> -1/6x = -9/20

=> x = -9/20 : (-1/6)

=> x = 27/10

\(\frac{1}{2}x+\frac{1}{5}=\frac{2}{3}x-\frac{1}{4}\)

\(\frac{1}{2}x=\frac{2}{3}x-\frac{1}{4}-\frac{1}{5}\)

\(\frac{1}{2}x=-\frac{9}{20}+\frac{2x}{3}\)

\(\frac{1}{2}x=-\frac{9}{20}+\frac{2x}{3}-\frac{2x}{3}\)

\(-\frac{x}{6}=-\frac{9}{20}\)

\(6\left(-\frac{x}{6}\right)=6\left(-\frac{9}{20}\right)\)

\(-x=-\frac{27}{10}\)

\(\Rightarrow x=-\frac{27}{10}\)

đặt x+2y=a ; xy=b

\(hpt\Leftrightarrow\hept{\begin{cases}a+b=6\\a^2-5b=6\end{cases}\Leftrightarrow\hept{\begin{cases}b=6-a\\a^2+5a-36=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}a=4;b=2\\a=-9;b=15\end{cases}}}\)

a/ \(\hept{\begin{cases}x+2y=4\\x.2y=4\end{cases}}\)=>x và 2y là nghiệm của pt \(X^2-4X+4=0\Rightarrow X=2\) hay x=2y=2 <=>x=2;y=1

b/\(\hept{\begin{cases}x+2y=-9\\x.2y=15\end{cases}}\)  tương tự TH a/ bạn tự giải nốt

\(=\frac{25}{4}-\left(x-\frac{1}{2}\right)^2\) xuống lớp 7 học đi nhé

GTLN \(-x^2\)+\(x\)+\(6\)=\(-\left(x^2-x-6\right)\)

=\(-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}-\frac{1}{4}-6\right)\)=\(-\left(x-\frac{1}{2}\right)^2-\frac{25}{4}\)

=\(\left(x-\frac{1}{2}\right)^2+\frac{25}{4}\)

Vì \(\left(x-\frac{1}{2}\right)^2\)\(\ge\)\(0\)Nên \(\left(x-\frac{1}{2}\right)^2+\frac{25}{4}\)\(\ge0\)

Vậy GTLN của biểu thức là \(\frac{25}{4}\)khi \(x=\frac{1}{2}\)

Tôi giải chi tiết đó nha :))