a, x+15=20-4x

=> x + 4x = 20 - 15

=> 5x = 5

=> x = 5 : 5

=> x = 1

b, 17-x=7-6x

=> -x + 6x = 7 - 17

=> 5x = - 10

=> x = -10 : 5

=> x = -2

c, 4.(x-5)-3.(x+7)= -19

=> 4x - 4.5 - 3.x + 3.7 = -19

=> 4x - 20 - 3x + 21 = -19

=> (4x-3x) + (21-20) = -19

=> 1x + 1 = -19

=> x = -19 - 1

=> x = -18

d, -7.(5-x)-2.(x-10)=15

=> -7x - (-7).5 - 2x - 2.10 = 15

=> -7x - (-35) - 2x - 20 = 15

=> -7x + 35 - 2x - 20 = 15

=> -7x+2x + (35-20) = 15

=> -5x + 15 = 15

=> -5x = 15 - 15

=> -5x = 0

=> x = 0 : (-5)

=> x = 0

e, -5.(2-x)+4.(x-3) = 10.x-15

=> -5x - (-5).2 + 4x - 4.3 = 10x - 15

=> -5x - (-10) + 4x - 12 = 10x - 15

=> -5x + 10 + 4x - 12 = 10x - 15

=> (-5x+4x) - (12-10) = 10x - 15

=> -1x - 2 = 10x - 15

=> -1x - 10x = -15 + 2

=> -11x = -13

=> x = \(\frac{-13}{-11}\)

a)A=4(x+11/8)^2 -153/16

Min A=-153/16 khi x=-11/8

b)B=3(x-1/3)^2 -4/3

Min B=-4/3 khi x=1/3

Bài 1:

a) \(A=4x^2+11x-2=\left(4x^2+11x+\dfrac{121}{16}\right)-\dfrac{153}{16}=\left(2x+\dfrac{11}{4}\right)^2-\dfrac{153}{16}\ge-\dfrac{153}{16}\)

\(minA=-\dfrac{153}{16}\Leftrightarrow x=-\dfrac{11}{8}\)

b) \(B=3x^2-2x-1=3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)-\dfrac{4}{3}=3\left(x-\dfrac{1}{3}\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\)

\(minB=-\dfrac{4}{3}\Leftrightarrow x=\dfrac{1}{3}\)

Bài 2:

a) \(A=-x^2+3x-1=-\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{5}{4}=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)

\(maxA=\dfrac{5}{4}\Leftrightarrow x=\dfrac{3}{2}\)

b) \(B=-x^2-4x+7=-\left(x^2+4x+4\right)+11=-\left(x+2\right)^2+11\le11\)

\(maxB=11\Leftrightarrow x=-2\)

Bài 1: 

a: Ta có: \(A=4x^2+11x-2\)

\(=4\left(x^2+\dfrac{11}{4}x-\dfrac{1}{2}\right)\)

\(=4\left(x^2+2\cdot x\cdot\dfrac{11}{8}+\dfrac{121}{64}-\dfrac{153}{64}\right)\)

\(=4\left(x+\dfrac{11}{8}\right)^2-\dfrac{153}{16}\ge-\dfrac{153}{16}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{11}{8}\)

b: Ta có: \(B=3x^2-2x-1\)

\(=3\left(x^2-\dfrac{2}{3}x-\dfrac{1}{3}\right)\)

\(=3\left(x^2-2\cdot x\cdot\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{4}{9}\right)\)

\(=3\left(x-\dfrac{1}{3}\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{3}\)

a: \(C=A+B=x^2-2y+xy+1+x^2+y-x^2y^2-1=2x^2-y+xy-x^2y^2\)

b: C=B-A

\(=x^2+y-x^2y^2-1-x^2+2y-xy-1\)

\(=-x^2y^2-2+3y-xy\)

\(a,C=A+B=x^2-2y+xy+1+x^2+y-x^2y^2-1\\ =2x^2-y+xy-x^2y^2\)

câu b đề khó hiểu quá

1) Để \(\left(x^2-1\right)\left(x^2-4\right)\left(x^2-7\right)\left(x^2-10\right)< 0\) thì phải có 1 số nhỏ hơn 0 hoặc 3 số nhỏ hơn 0

TH1 : có 1 số nhỏ hơn 0

Vì \(x^2-1>x^2-4>x^2-7>x^2-10\)

Nên \(\hept{\begin{cases}x^2-1;x^2-4;x^2-7>0\\x^2-10< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2-7>0\\x^2-10< 0\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2>7\\x^2< 10\end{cases}\Leftrightarrow7< x^2< 10\Rightarrow x^2=9\Rightarrow x=\pm3}\)

TH2: 3 số nhỏ hơn 0

Vì \(x^2-1>x^2-4>x^2-7>x^2-10\)

Nên \(\hept{\begin{cases}x^2-1>0\\x^2-4;x^2-7;x^2-10< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2-1>0\\x^2-4< 0\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2>1\\x^2< 4\end{cases}\Rightarrow1< x^2< 4}\) (loại vì x là số nguyên)

Vậy \(x=\pm3\)

2) \(A=\left|x-a\right|+\left|x-b\right|+\left|x-c\right|+\left|x-d\right|\)

\(=\left|x-a\right|+\left|x-d\right|+\left|x-c\right|+\left|x-b\right|\)

\(=\left|x-a\right|+\left|d-x\right|+\left|x-c\right|+\left|b-x\right|\)

\(\ge\left|x-a+d-x\right|+\left|x-c+b-x\right|=\left|d-a\right|+\left|b-c\right|=c+d-a-b\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-a\right)\left(d-x\right)\ge0\\\left(x-c\right)\left(b-x\right)\ge0\end{cases}\Rightarrow b\le x\le c}\)

Vậy GTNN của A là \(c+d-a-b\) tại \(b\le x\le c\)

z=-6

y=9

x=-7

x=-7

y=9

z=-6

a)\(A=1+x+x^2+x^3+..........+x^{2012}\)

+)Thay x=1 vào biểu thức đc:

\(A=1+1+1^2+1^3+..............+1^{2012}\)

               Có 2013 số hạng

\(\Rightarrow A=1.2013=2013\)

b)\(B=1-x+x^2-x^3+..............-x^{2011}\)

\(\Rightarrow B=\left(1-x\right)+\left(x^2-x^3\right)+............+\left(x^{2010}-x^{2011}\right)\)

+)Thay x=1 vào biểu thức được:

\(B=\left(1-1\right)+\left(1^2-1^3\right)+...........+\left(1^{2010}-1^{2011}\right)\)

\(\Rightarrow B=0+0+......................+0=0\)

+)\(C=A+B\Rightarrow C=2013+0\Rightarrow C=2013\)

Vậy C=2013

Chúc bn học tốt

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

\(x=\frac{5}{12}\)

Vậy...

b)\(x=\frac{5}{7}+\frac{2}{5}\)

\(x=\frac{39}{35}\)

Vậy...

c)\(-x=\frac{6}{7}+\frac{2}{3}\)

\(-x=\frac{32}{21}\)

\(x=\frac{-32}{21}\)

Vậy...

d)\(-x=\frac{1}{3}-\frac{4}{7}\)

\(-x=\frac{-5}{21}\)

\(x=\frac{5}{21}\)

Vậy...

tk mk nhoaa bn