bài vách ngọc ngà và bài cà phê ko đường

a) \(x(x+y)-y(x+y)\)

Ta có:

\(= (x-y)(x+y)\)

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

Tính \(x,y\):

\(x=\left(-\dfrac12\right)^5:\left(\dfrac12\right)^4\)

\(=-\dfrac1{32}:\dfrac1{16}=-\dfrac12\)

\(y=8^2:(-2)^5\)

\(=64:(-32)=-2\)

Thay vào biểu thức:

\(A=\left(-\dfrac12\right)^2-(-2)^2\)

\(=\dfrac14-4\)

\(=-\dfrac{15}{4}\)

b) \((x-y)(x^2+xy+y^2)-(x+y)(x^2-y^2)\), biết \(x-y=0\).

Ta có:

\(x-y=0\Rightarrow x=y\).

\(B=0\cdot(x^2+xy+y^2)-(x+y)(x-y)(x+y)\)

\(=0-0\)

\(=0\).

c) \(x^3(x^2-y^2)+y^2(x^3-y^3)\)

Biết \(x=16^5:8^5:(-2)^4\) và \(|y|=1\).

Tính \(x\):

\(x=\dfrac{16^5}{8^5}:16\)

\(=2^5:16=32:16=2\).

Vì \(|y|=1\) nên \(y=1\) hoặc \(y=-1\).

Ta rút gọn biểu thức:

\(C=x^5-x^3y^2+x^3y^2-y^5\)

\(=x^5-y^5\).

- Nếu \(y=1\): \(C=2^5-1^5=31\).

- Nếu \(y=-1\): \(C=2^5-(-1)^5=33\).

Vậy \(C=31\) hoặc \(C=33\).

d) Câu d bạn ghi các giá trị:

- \(x=y=0\)

- \(x=y=1\)

- \(x=\dfrac12,\ y=-\dfrac32\)

- \(x=\sqrt4,\ y=\sqrt9\)

Nhưng chưa có biểu thức cần thay số. Bạn gửi lại biểu thức của câu d nhé.

a)<=>

A,=(x+y)(x-y)=x^2-y^2

x=(-1/2)^5:(1/2)^4=-1/2

x^2=1/4

y=8^2/(-2)^5=-2

y^2=4

A=1/4-4=-15/4

a, ( 2x - 3 )²- (2x + 1)² = -3

4x²-12x+9-4x²+4x-1=-3

-8x-1=-3

-8x=-2

x=\(\frac{1}{4}\)

b, (5x - 1) ² - (5x + 4)(5x - 4) = 7

25x²-10x+1-25x²+16=7

-10x+17=7

-10x=-10

x=1

c, ( x- 5)² + (x-3)(x+3) - 2(x + 1)²=0

x²-10x+25+x²-9-2x²-4x-2=0

-14x+14=0

-14(x-1)=0

=>x-1=0

x=1

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

\(\Leftrightarrow4x^2-12x+9-4x^2-4x-1=-3\)

\(\Leftrightarrow-16x+8=-3\)

\(\Leftrightarrow-16x=-11\)

\(\Leftrightarrow x=\frac{11}{16}\)

b)\(\left(5x-1\right)^2-\left(5x+4\right)\left(5x-4\right)=7\)

\(\Leftrightarrow25x^2-10x+1-25x^2+16=7\)

\(\Leftrightarrow-10x+17=7\)

\(\Leftrightarrow-10x=-10\)

\(\Leftrightarrow x=1\)

c)\(\left(x-5\right)^2+\left(x-3\right)\left(x+3\right)-2\left(x+1\right)^2=0\)

\(\Leftrightarrow x^2-10x+25+x^2-9-2\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow2x^2-10x-16-2x^2-4x-2=0\)

\(\Leftrightarrow-14x-18=0\)

\(\Leftrightarrow-14x=18\)

\(\Leftrightarrow x=-\frac{9}{7}\)

#H

\(a)\)

\(21\left(x+3\right)^3:\left(3x+9\right)^2\)

\(=[21\left(x+3\right)^3]:[3^2\left(x+3\right)^2]\)

\(=7\left(x+3\right):3\)

Thay vào ta được: \(7.\frac{\left(-6+3\right)}{3}=7.\left(-3\right):3=-7\)

\(b)\)

Thay vào ta được:

\(\left(2.2^2-5.2+3\right)^4:[\left(2.2-3\right)^3:\left(2-1\right)^2]\)

\(=\left(2.4-10+3\right)^4:[\left(4-3\right)^31^2]\)

\(=1^4:\left(1^3.1\right)\)

\(=1:1\)

\(=1\)

\(c)\)

Thay vào ta được:

\(36.10^4.7^3:\left(-6.10^3.7^2\right)\)

\(=-6.10.7\)

\(=-420\)

5/ (x² – 4) + (x – 2)(4 – 2x) = 0

⇔(x-2)(x+2)+(x – 2)(4 – 2x)=0

⇔(x-2)(x+2+4-2x)=0

⇔(x-2)(6-x)=0

⇔\(\left[{}\begin{matrix}x-2=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

6/ x(2x – 7) – 4x + 14 = 0

⇔2x²-11x+14=0

⇔(x-\(\frac{7}{2}\))(x-2)=0

⇔\(\left[{}\begin{matrix}x-\frac{7}{2}=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=2\end{matrix}\right.\)

7/ x² – x – (3x–3)= 0

⇔x²-4x+3=0

⇔(x-3)(x-1)=0

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

8/ (x² – 2x + 1) – 4 = 0

⇔(x-1)²-4=0

⇔(x-1-4)(x-1+4)=0

⇔(x-5)(x+3)=0

⇔\(\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

9/ 4x² + 4x + 1 = x²

⇔3x²+4x+1=0

⇔(3x+1)(x+1)=0

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

10/ x² – x = - 2x + 2

⇔3x²-x-2=0 (chuyển vế)

⇔(3x+2)(x-1)=0

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

11/ x² – 5x + 6 = 0

⇔x²-3x-2x+6=0

⇔x(x-3)-2(x-3)=0

⇔(x-3)(x-2)=0

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

Mình làm bài khá tắt nên có gì không hiểu bạn cứ hỏi mình nha!

5/(x²-4)+(x-2)(4-2x)=0⇔x²-4+4x-2x²-8+4x=0

⇔-x²+8x-12=0⇔-x²+2x+6x-12=0⇔-x(x-2)-6(x-2)=0⇔(-x-6)(x-2)=0

⇔-x-6=0 hoặc x-2=0⇔x=-6 hoặc x=2

1, \(3x\left(x-7\right)+2x-14=0\)

\(\Rightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)

\(\Rightarrow\left(x-7\right)\left(3x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=7\\x=\frac{-2}{3}\end{cases}}\)

2, \(x^3+3x^2-\left(x+3\right)=0\)

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

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

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

\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)

3, \(15x-5+6x^2-2x=0\)

\(\Rightarrow\left(15x-5\right)+\left(6x^2-2x\right)=0\)

\(\Rightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)

\(\Rightarrow\left(3x-1\right)\left(5+2x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{-5}{2}\end{cases}}\)

4, \(5x-2-25x^2+10x=0\)

\(\Rightarrow\left(5x-25x^2\right)-\left(2-10x\right)=0\)

\(\Rightarrow5x\left(1-5x\right)-2\left(1-5x\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}1-5x=0\\5x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{2}{5}\end{cases}}\)