A) gtln B =5 khi x =2

\(\Leftrightarrow B=-\left(x^2-4x-1\right)\)

\(\Leftrightarrow B=-\left(x^2-4x+4-5\right)\)

\(\Leftrightarrow B=-\left(x-2\right)^2+5\)

Ta có \(\left(x-2\right)^2\ge0\)với mọi x

  \(\Leftrightarrow-\left(x-2\right)^2\le0\)

\(\Leftrightarrow-\left(x-2\right)^2+5\le0+5\)

hay               \(B\)            \(\le5\)

Dấu "=" xảy ra khi \(\left(x-2\right)^2=0\)

                     .  \(\Leftrightarrow x-2\)\(=0\)

                       \(\Leftrightarrow\)\(x\)      \(=2\)

Vậy min B=5 tại x=2

lam cau b ay ban cau a minh lam roi

1) a) \(\left(3x-1\right)\left(9x^2+3x+1\right)-4x\left(x-5\right)\)

\(=27x^3+9x^2+3x-9x^2-3x-1-4x^2+20x\)

\(=27x^3+\left(9x^2-9x^2-4x^2\right)+\left(3x-3x+20x\right)+\left(-1\right)\)

\(=27x^3-4x^2+20x-1\)

b)\(\left(7x+2\right)\left(3-4x\right)-\left(x+3\right)\left(x^2-3x+9\right)\)

\(=21x-28x^2+6-8x-x^3+3x^2-9x-3x^2+9x-27\)

\(=\left(21x-8x-9x+9x\right)+\left(-28x^2+3x^2-3x^2\right)\)\(+\left(6-27\right)\)\(+\left(-x^3\right)\)

\(=13x-28x^2-21-x^3\)

c)\(\left(4x+3\right)\left(4x-3\right)-\left(2-x\right)\left(4+2x+x^2\right)\)

\(=16x^2-12x+12x-9-8-4x-2x^2+4x+2x^2+x^3\)

\(=\left(16x^2-2x^2+2x^2\right)+\left(-12x+12x-4x+4x\right)\)\(+\left(-9-8\right)\)\(+x^3\)

\(=16x^2-17+x^3\)

d)\(\left(3x-8\right)\left(-5x+6\right)-\left(4x+1\right)\left(3x-2\right)\)

\(=-15x^2+18x+40x-48-12x^2+8x-3x+2\)

\(=\left(-15x^2-12x^2\right)+\left(18x+40x+8x-3x\right)\)\(+\left(-48+2\right)\)

\(=-27x^2+63x-46\)

e)\(\left(3x-6\right)4x-2x\left(3x+5\right)-4x^2\)

\(=12x^2-24x-6x^2-10x-4x^2\)

\(=\left(12x^2-6x^2-4x^2\right)+\left(-24x-10x\right)\)

\(=2x^2-34x\)

f)\(\left(5x-6\right)\left(6x-5\right)-x\left(3x+10\right)\)

\(=30x^2-25x-36x+30-3x^2-10x\)

\(=\left(30x^2-3x^2\right)+\left(-25x-36x-10x\right)+30\)

\(=27x^2-71x+30\)

2) a)\(x\left(x+3\right)-x^2=6\)

\(\Rightarrow x^2+3x-x^2=6\)

\(\Rightarrow\left(x^2-x^2\right)+3x=6\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

Vậy x=2

b) \(2x\left(x-5\right)+x\left(-2x-1\right)=6\)

\(\Rightarrow2x^2-10x-2x^2-x=6\)

\(\Rightarrow\left(2x^2-2x^2\right)+\left(-10x-x\right)=6\)

\(\Rightarrow-11x=6\)

\(\Rightarrow x=-\dfrac{6}{11}\)

\(\)Vậy \(x=-\dfrac{6}{11}\)

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

\(\Rightarrow x^2+5x-x^2+2x-x+2=7\)

\(\Rightarrow\left(x^2-x^2\right)+\left(5x+2x-x\right)=7-2\)

\(\Rightarrow6x=5\)

\(\Rightarrow x=\dfrac{5}{6}\)

Vậy x=\(\dfrac{5}{6}\)

d)\(\left(3x+4\right)\left(6x-3\right)-\left(2x+1\right)\left(9x-2\right)=10\)

\(\Rightarrow18x^2-9x+24x-12-18x^2+4x-9x+2=10\)

\(\Rightarrow\left(18x^2-18x^2\right)+\left(-9x+24x+4x-9x\right)+\left(-12+2\right)=10\)

\(\Rightarrow10x-10=10\)

\(\Rightarrow10x=20\)

\(\Rightarrow x=2\)

Vậy x=2

b: \(\left(2x+1\right)^2=25\)

=>\(\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

c: \(\left(1-3x\right)^3=64\)

=>\(\left(1-3x\right)^3=4^3\)

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1

d: \(\left(4-x\right)^3=-27\)

=>\(\left(4-x\right)^3=\left(-3\right)^3\)

=>4-x=-3

=>x=4+3=7

e: \(x^2-5x=0\)

=>\(x\left(x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

a,\(\left(x-7\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)

Vậy...

b,\(\left(x-5\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-9=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=9\end{matrix}\right.\)

Vậy...

c,\(\left(x-5\right)\left(x^2-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2-9=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)

Vậy...

Câu d bạn viết lại đề nhé

Theo mink làm như zậy nè

a) (x-7)(x+7)=0

Ta xét 2 TH

TH1 nếu x-7 =0 thì x=7

\(\Rightarrow\)x+7 ta tìm đc x là x\(\in\)R

TH2 nêu x+7=0 thì x=-7

\(\Rightarrow\)x-7 ta tìm được x là \(x\in R\)

Câu b,c,d tương tự nhé

a, (x-7).(x+7) = 0
\(\Rightarrow\left[{}\begin{matrix}x-7=0\\x+7=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0+7\\x=0-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)

b,(x-5).(x-9)=0
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+5\\x=0+9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=9\end{matrix}\right.\)

2)  \(x^3-6x^2+11x-6=0\)

\(\Leftrightarrow\)\(x^3-3x^2-3x^2+9x+2x-6=0\)

\(\Leftrightarrow\)\(\left(x-3\right)\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow\)\(\left(x-3\right)\left(x-2\right)\left(x-1\right)=0\)

bn giải tiếp nha

3)   \(x^3-4x^2+x+6=0\)

\(\Leftrightarrow\)\(x^3-3x^2-x^2+3x-2x+6=0\)

\(\Leftrightarrow\)\(\left(x-3\right)\left(x^2-x-2\right)=0\)

\(\Leftrightarrow\)\(\left(x-3\right)\left(x-2\right)\left(x+1\right)=0\)

lm tiếp nha

4)  \(x^3-3x^2+4=0\)

\(\Leftrightarrow\)\(x^3+x^2-4x^2-4x+4x+4=0\)

\(\Leftrightarrow\)\(\left(x+1\right)\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\)\( \left(x+1\right)\left(x-2\right)^2=0\)

lm tiếp nha

Mk làm mẫu 1 bài cho nha !

1. <=> (x^3-x^2)+(5x^2-5x)+(6x-6) = 0

<=> (x-1).(x^2+5x+6) = 0

<=> (x-1).[(x^2+2x)+(3x+6)] = 0

<=> (x-1).(x+2).(x+3) = 0

<=> x-1=0 hoặc x+2=0 hoặc x+3=0

<=> x=1 hoặc x=-2 hoặc x=-3

Vậy ..............

Tk mk nha

2.  x3−6x2+11x−6=0

⇔x3−3x2−3x2+9x+2x−6=0

⇔(x−3)(x2−3x+2)=0

⇔(x−3)(x−2)(x−1)=0

bn giải tiếp nha

3)   x3−4x2+x+6=0

⇔x3−3x2−x2+3x−2x+6=0

⇔(x−3)(x2−x−2)=0

⇔(x−3)(x−2)(x+1)=0

lm tiếp nha

4)  x3−3x2+4=0

⇔x3+x2−4x2−4x+4x+4=0

⇔(x+1)(x2−4x+4)=0

⇔(x+1)(x−2)2=0

lm tiếp nha

a)    bạn nhóm 2 cái cuối thành 1 nhóm,  2 cái ở giữa thành 1 nhóm,  rồi đặt ẩn phụ là ra

         \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)

\(\Leftrightarrow\)\(\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0\)

Đặt  \(x^2+3x=t\)   ta có:

                  \(t\left(t+2\right)-24=0\)

\(\Leftrightarrow\)\(t^2+2t-24=0\)

\(\Leftrightarrow\)\(\left(t-4\right)\left(t+6\right)=0\)

đến đây bn thay trở lại rồi tìm nghiệm nhé

a, x(x+3)(x+1)(x+2)-24=0

=> (x^2+3x)(x^2+3x+2)-24=0

đặt x^2+3x=a

ta có : a(a+2)-24=0

=> a^2+2a-24=0 => \(\orbr{\begin{cases}a=4\\a=-6\end{cases}}\) giải ra ta được x^2+3x=4 hay \(\orbr{\begin{cases}x=-4\\x=1\end{cases}}\)

và x^2+3x=-6 => vô nghiệm vậy x=-4 hoặc x=1

b, 5x(2-3x)-(4-6x)=0

=> 5x(2-3x)-2(2-3x)=0

=>(5x-2)(2-3x)=0 

=>\(\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{2}{5}\end{cases}}\)

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

A, -5(x-7)-7(x+3)=8 <=> -5x+35-7x+21=8<=>-12x=-48<=>x=4

B, 6(x+1)-5(x-2)=8<=>6x+6-5x+10=8<=>x=-8

C, 3x+4x+5x+6x=30<=>18x=30<=>x=\(\dfrac{5}{3}\)

D, (x+1)+(x+2)+(x+3)+...+(x+10)=144

<=> (x+x+x+x+x+x+x+x+x+x)+(1+2+3+4+5+6+7+8+9+10)=144

<=>10x+55=144<=>10x=89<=>x=\(\dfrac{89}{10}\)