a: Ta có: \(\left(x-2\right)^3-x\left(x+1\right)\left(x-1\right)+6x^2=5\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2=5\)

\(\Leftrightarrow13x=13\)

hay x=1

\((x-2)^3-(x+5)(x^2-5x+25)+6x^2=11\\\Leftrightarrow (x-2)^3-(x+5)(x^2-5.x+5^2)+6x^2=11 \\\Leftrightarrow x^3-6x^2+12x-8 -(x^3+5^3)+6x^2-11=0 \\\Leftrightarrow 12x-144=0 \\\Leftrightarrow x=12\)

Vậy \(x=12\).

(x−2)3−(x+5)(x2−5x+25)+6x2=11

=>(x−2)3−(x+5)(x2−5.x+52)+6x2=11

=>x3−6x2+12x−8−(x3+53)+6x2−11=0

=>12x−144=0

=>x=12(x−2)3−(x+5)(x2−5x+25)+6x2=11

=>(x−2)3−(x+5)(x2−5.x+52)+6x2=11

=>x3−6x2+12x−8−(x3+53)+6x2−11=0

=>12x−144=0

=>x=12

Vậy x=12x=12.

cho tôi đúng đi

\(\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)+6x^2=11\)

=>\(x^3-6x^2+12x-8-\left(x^3+125\right)+6x^2=11\)

=>\(x^3+12x-8-x^3-125=11\)

=>12x-133=11

=>12x=144

=>\(x=\dfrac{144}{12}=12\)

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

\(\Rightarrow x^3+3x^2+9x-3x^2-9x-27-x\left(x^2-16\right)=5\)

\(\Rightarrow x^3-27-x^3-16x=5\)

\(\Rightarrow-16x-27=5\)

\(\Rightarrow-16x=32\Rightarrow x=-2\)

b, \(\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)+6x^2=11\)

\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-5x^2+25x+5x^2-25x+125\right)+6x^2=11\)

\(\Rightarrow x^3-6x^2+12x-8-x^3-125+6x^2=11\)

\(\Rightarrow12x-133=11\Rightarrow12x=144\Rightarrow x=12\)

Chúc bạn học tốt!!!

a)

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

\(\Rightarrow x^3-3^3-x.\left(x^2-16\right)=5\)

\(\Rightarrow x^3-27-x^3+16.x=5\)

\(\Rightarrow16x-27=5\)

\(\Rightarrow16x=32\)

\(\Rightarrow x=2\)

Vậy x = 2

b)

\(\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)+6x^2=11\)

\(\Rightarrow x^3-6x^2+12x-8-x^3-125+6x^2=11\)

\(\Rightarrow12x-133=11\)

\(\Rightarrow12x=144\)

\(\Rightarrow x=12\)

Vậy x = 12

nhấn vào đúng 0 sẽ có câu trả lời

+)   (5x-1). (2x+3)-3. (3x-1)=0

10x^2+15x-2x-3 - 9x+3=0

10x^2 +8x=0

2x(5x+4)=0

=> x=0 hoặc x= -4/5

+)    x^3 (2x-3)-x^2 (4x^2-6x+2)=0

2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0

-2x^4 + 3x^3-2x^2=0

x^2(-2x^2+x-2)=0

-2x^2(x-1)^2=0

=> x=0 hoặc x=1

+)   x (x-1)-x^2+2x=5

x^2 -x -x^2+2x=5

x=5

+)     8 (x-2)-2 (3x-4)=25

8x - 16-6x+8=25

2x=33

x=33/2

a. 2x\(^2\)-8=0

2x\(^2\)=8

x\(^2\)=4

x=2

b.3x\(^3\)-5x=0

x(3x\(^2\)-5)=0

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

c.x+3x-4=0\(^{\left(\cdot\right)}\)

đặt t=x\(^2\) (t>0)

ta có pt: t\(^2\)+3t-4=0 \(^{\left(1\right)}\)

thấy có a+b+c=1+3+(-4)=0 nên pt\(^{\left(1\right)}\) có 2 nghiệm

t\(_1\)=1; t\(_2\)=\(\dfrac{c}{a}\)=-4

khi t\(_1\)=1 thì x\(^2\)=1 ⇒x=\(^+_-\)1

khi t\(_2\)=-4 thì x\(^2\)=-4 ⇒ x=\(^+_-\)2

vậy pt đã cho có 4 nghiệm x=\(^+_-\)1; x=\(^+_-\)2

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

thấy có a+b+c= 3+6+(-9)=0 nên pt có 2 nghiệm

x\(_1\)=1; x\(_2\)=\(\dfrac{c}{a}=\dfrac{-9}{3}=-3\)

e. \(\dfrac{x+2}{x-5}+3=\dfrac{6}{2-x}\)  (ĐK: x#5; x#2 )

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

⇒2x - x\(^2\) + 4 - 2x + 6x - 6x\(^2\) + 12 - 6x - 6x +30 = 0

⇔-7x\(^2\) - 6x + 46=0

Δ'=b'\(^2\)-ac = (-3)\(^2\) - (-7)\(\times\)46= 9+53 = 62>0

\(\sqrt{\Delta'}=\sqrt{62}\)

vậy pt có 2 nghiệm phân biệt

x\(_1\)=\(\dfrac{-b'+\sqrt{\Delta'}}{a}=\dfrac{3+\sqrt{62}}{-7}\)

x\(_2\)=\(\dfrac{-b'-\sqrt{\Delta'}}{a}=\dfrac{3-\sqrt{62}}{-7}\)

vậy pt đã cho có 2 nghiệm x\(_1\)=.....;x\(_2\)=......

câu g làm tương tự câu c

tìm x nha

+(x-3)²-x²=11

x²-6x+9-x²=11

-6x+9=11

-6x=2

x=2:-6

x=-1/3

(6x-3)²-36x(x-1)=40

36x²-36x+9-36x²+36x=40

9=40

=> đề sai

(2-x)²-7(x²+11)=0

4-4x+x²-7x²-77=0

-73-4x-6x²

ht bt