e, (x-1)(x² + x + 1)-x(x+2)(x-2) = 5

x(x² +x + 1 ) - (x² + x +1 )- [ x (x² - 4)] = 5

x³ +x² +x - x² - x - 1 - x³ +4x = 5

4x - 1 = 5

4x = 6

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

f, (x-1)³ - (x+3)(x² - 3x +9 ) +3(x² - 4) = 2

x - 3x² +3x - 1 - [( x³ - 3x² + 9x) + (3x² - 9x +27)] = 2

x³ - 3x² + 3x - 1 -x³ +3x² -9x - 3x² +9x - 27 +3x² - 12 = 2

3x - 1 - 27 - 12 = 2

3x = 42

x = 14

muốn tao trả lờ cho ko , mai đến lớp nhá

làm nốt

d) (2x-1)(3x+2)(3-x)

=(6x²+x-2)(3-x)

=-6x³+17x²+5x-6

e) (x+3)(x²+3x-5)

=x³+6x²+4x-15

f) (xy-2)(x³-2x-6)

=x⁴y-2x³-2x²y-6xy+4x+12

g) (5x³-x2+2x-3)(4x²-x+2)

=20x⁵-9x⁴+19x³-16x²+7x-6

Bài 1:

a) (x-2)(x2+3x+4)

=x(5x+4)-2(5x+4)

= 5x²+4x-10x-8

=5x²-6x-8

Bài 2: a) \(3x^3-3x=0\Leftrightarrow3x\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

b) \(x^2-x+\frac{1}{4}=0\Leftrightarrow x^2-2.\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)

Bạn làm bài kiểm tra hả sao nhiều bài tek. Mk làm mất khá nhiều tg luôn đó

Có một số câu thì mình không làm được. Mong bạn thông cảm!!!

Ai giúp mình với ạh

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

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

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

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

Suy ra \(x\) =2,685673906

2)

a) \(3x^3-3x=0\)

\(\Leftrightarrow3x\left(x^2-1\right)=0\)

\(\Leftrightarrow3x\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-1=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

Vậy x=0 ; x=-1 ; x=1

b) \(x^2-x+\dfrac{1}{4}=0\)

\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow x-\dfrac{1}{2}=0\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

1)

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

\(\Leftrightarrow x^3+3x^2+4x-2x^2-6x-8\)

\(\Leftrightarrow x^3+x^2-2x-8\)

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

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

\(=3x^2-x^3+2x-8\)

c) \(\left(x^2-1\right)\left(x^2+2x\right)\)

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

d) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)

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

\(=18x^2+12x-9x-6-6x^3-4x^2+3x^2+2x\)

\(=17x^2+5x-6-6x^3\)

a) \(x^3+8=x^3+2^3=\left(x+2\right)\left(x^2-2x+4\right)\)

b) \(27-8y^3=3^3-\left(2y\right)^3=\left(3-2y\right)\left(9+6y+4y^2\right)\)

c) \(\left(x-2\right)^3=\left(x-2\right)\left(x-2\right)\left(x-2\right)\)

d) \(\left(x+1\right)^3=\left(x+1\right)\left(x+1\right)\left(x+1\right)\)

P/s: Câu e, f đã biến đổi rồi mà bạn

a) Ta có: \(x^3+8\)

\(=x^3+2^3\)

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

b) Ta có: \(27-8y^3\)

\(=3^3-\left(2y\right)^3\)

\(=\left(3-2y\right)\left(9+6y+4y^2\right)\)

a) x vô nghiệm

b)<=>(x²-3x+3)(x²-2x+3)-2x²=(x-3)(x-1)(x²-x+3)

=>(x-3)(x-1)(x²-x+3)=0

TH1:x-3=0

=>X=3

TH2:x-1=0

=>x=1

TH3:x²-x+3=0

<=>(-1)²-4(1.3)=-11

vì -11<0

=>x=1 hoặc 3

bạn tự tiếp làm đi dễ mà

a) Ta có: \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)

\(\Leftrightarrow\frac{\left(2x+1\right)^2\cdot3}{15}-\frac{5\left(x-1\right)^2}{15}-\frac{7x^2-14x-5}{15}=0\)

\(\Leftrightarrow3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)-7x^2+14x+5=0\)

\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)

\(\Leftrightarrow36x+3=0\)

\(\Leftrightarrow36x=-3\)

\(\Leftrightarrow x=\frac{-3}{36}\)

Vậy: \(x=\frac{-3}{36}\)

b) Ta có: \(\frac{201-x}{99}+\frac{203-x}{97}=\frac{205-x}{95}+3=0\)

\(\Leftrightarrow\frac{201-x}{99}+\frac{203-x}{97}-\frac{205-x}{95}-3=0\)

\(\Leftrightarrow\left(\frac{201-x}{99}+1\right)+\left(\frac{203-x}{97}+1\right)+\left(\frac{205-x}{95}+1\right)=0\)

\(\Leftrightarrow\frac{201-x+99}{99}+\frac{203-x+97}{97}+\frac{205-x+95}{95}=0\)

\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)

\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\ne0\)

nên 300-x=0

\(\Leftrightarrow x=300\)

Vậy: x=300

c) Ta có: \(x^3+x^2+x+1=0\)

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

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

Ta có: \(x^2\ge0\forall x\)

\(\Rightarrow x^2+1\ge1\ne0\forall x\)(2)

Từ (1) và (2) suy ra x+1=0

hay x=-1

Vậy: x=-1

d) Ta có: \(\left(x-1\right)x\left(x+1\right)\left(x+2\right)=24\)

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

Đặt \(x^2+x-1=t\)

\(\Leftrightarrow\left(t+1\right)\left(t-1\right)=24\)

\(\Leftrightarrow t^2-1-24=0\)

\(\Leftrightarrow t^2-25=0\)

\(\Leftrightarrow\left(t-5\right)\left(t+5\right)=0\)

\(\Leftrightarrow\left(x^2+x-1-5\right)\left(x^2+x-1+5\right)=0\)

\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+x+4\right)=0\)

\(\Leftrightarrow\left(x^2+3x-2x-6\right)\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{15}{4}\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-2\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{15}{4}=0\right]\)(3)

Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}\ne0\forall x\)(4)

Từ (3) và (4) suy ra

\(\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

Vậy: \(x\in\left\{-3;2\right\}\)

e) Ta có: \(\left(5x-3\right)-\left(4x-7\right)=0\)

\(\Leftrightarrow5x-3-4x+7=0\)

\(\Leftrightarrow x+4=0\)

\(\Leftrightarrow x=-4\)

Vậy: x=-4

f) Ta có: \(3x^2+2x-1=0\)

\(\Leftrightarrow3x^2+3x-x-1=0\)

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{1}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{-1;\frac{1}{3}\right\}\)

g) Ta có: \(x^2+6x-16=0\)

\(\Leftrightarrow x^2-2x+8x-16=0\)

\(\Leftrightarrow x\left(x-2\right)+8\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)

Vậy: \(x\in\left\{2;-8\right\}\)

h) Ta có: \(x^2+3x-10=0\)

\(\Leftrightarrow x^2+5x-2x-10=0\)

\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)

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

Vậy: \(x\in\left\{-5;2\right\}\)

i) Ta có: \(x^2+x-2=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{1;-2\right\}\)

k) Ta có: \(3x^2+7x+2=0\)

\(\Leftrightarrow3x^2+6x+x+2=0\)

\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{-1}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{-2;\frac{-1}{3}\right\}\)

l) Ta có: \(4x^2-12x+5=0\)

\(\Leftrightarrow4x^2-2x-10x+5=0\)

\(\Leftrightarrow2x\left(2x-1\right)-5\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{5}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{1}{2};\frac{5}{2}\right\}\)