Bài 3: 

Xét ΔIAB có 

\(\widehat{AIB}+\widehat{IAB}+\widehat{IBA}=180^0\)

\(\Leftrightarrow\widehat{IAB}+\widehat{IBA}=115^0\)

hay \(\widehat{DAB}+\widehat{ABC}=230^0\)

Xét tứ giác ABCD có 

\(\widehat{D}+\widehat{C}+\widehat{DAB}+\widehat{CBA}=360^0\)

\(\Leftrightarrow\widehat{D}+\widehat{C}=150^0\)

mà \(\widehat{C}-\widehat{D}=10^0\)

nên \(2\cdot\widehat{C}=160^0\)

\(\Leftrightarrow\widehat{C}=80^0\)

\(\Leftrightarrow\widehat{D}=70^0\)

1: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)

=>\(x^3-9x^2+27x-27-\left(x^3-27\right)+9\left(x^2+2x+1\right)=15\)

=>\(-9x^2+27x+9x^2+18x+9=15\)

=>45x=6

=>\(x=\frac{6}{45}=\frac{2}{15}\)

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

=>\(x\left(x^2-25\right)-\left(x^3+8\right)=3\)

=>\(x^3-25x-x^3-8=3\)

=>-25x=11

=>\(x=-\frac{11}{25}\)

3: \(\left(x+4\right)\left(x^2-4x+16\right)-x\left(x-5\right)\left(x+5\right)=264\)

=>\(x^3+64-x\left(x^2-25\right)=264\)

=>\(x^3+64-x^3+25x=264\)

=>25x=200

=>x=8

4: \(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)+6\left(x-2\right)\left(x+2\right)=60\)

=>\(x^3-6x^2+12x-8-\left(x^3-8\right)+6\left(x^2-4\right)=60\)

=>\(-6x^2+12x+6x^2-24=60\)

=>12x-24=60

=>12x=84

=>x=7

5: \(\left(x+3\right)^4-\left(x-3\right)^4-24x^3=108\)

=>\(\left\lbrack\left(x+3\right)^2-\left(x-3\right)^2\right\rbrack\left\lbrack\left(x+3\right)^2+\left(x-3\right)^2\right\rbrack-24x^3=108\)

=>\(\left(x^2+6x+9-x^2+6x-9\right)\left(x^2+6x+9+x^2-6x+9\right)-24x^3=108\)

=>\(12x\left(2x^3+18\right)-24x^3=108\)

=>\(24x^3+216x-24x^3=108\)

=>216x=108

=>\(x=\frac{108}{216}=\frac12\)

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

=>\(25x^2-10x+1-\left(25x^2-16\right)=7\)

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

=>-10x=7-17=-10

=>x=1

8: \(\left(4x+1\right)^2-\left(2x+3\right)^2+5\left(x+2\right)^2+3\left(x-2\right)\left(x+2\right)=500\)

=>\(16x^2+8x+1-\left(4x^2+12x+9\right)+5\left(x^2+4x+4\right)+3\left(x^2-4\right)\) =500

=>\(16x^2+8x+1-4x^2-12x-9+5x^2+20x+20+3x^2-12=500\)

=>\(20x^2+16x-500=0\)

=>\(x^2+\frac45x-25=0\)

=>\(x^2+2\cdot x\cdot\frac25+\frac{4}{25}-25-\frac{4}{25}=0\)

=>\(\left(x+\frac25\right)^2=25+\frac{4}{25}=\frac{629}{25}\)

=>\(\left[\begin{array}{l}x+\frac25=\frac{\sqrt{629}}{5}\\ x+\frac25=-\frac{\sqrt{629}}{5}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt{629}-2}{5}\left(nhận\right)\\ x=\frac{-\sqrt{629}-2}{5}\left(nhận\right)\end{array}\right.\)

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

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

=>4x-27=1

=>4x=28

=>x=7

10: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

=>\(x^3+3x^2+3x+1-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)=-10\)

=>\(x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-10\)

=>12x-6=-10

=>12x=-4

=>x=-4/12=-1/3

lớp lơn hơn có thể giải bài này mà đâu nhất thiết phải là lớp 5 

Bài giải 1

Diện tích thửa ruộng hình chữ nhật (1):

40 ⨯ (40 + 30) = 2800 (m²)

Diện tích thửa ruộng hình chữ nhật (2):

40 ⨯ (60,5 – 40) = 820 (m²)

Diện tích thửa ruộng hình bên là:

2800 + 820 = 3620 (m²)

Đáp số: 3620m²

Bài giải 2

Diện tích mảnh đất hình chữ nhật (1):

50 ⨯ 20,5 = 1025 (m²)

Diện tích mảnh đất hình chữ nhật (2):

10 ⨯ 40,5 = 405 (m²)

Diện tích mảnh đất hình bên là:

1025 + 405 = 1430 (m²)

Đáp số: 1430m²

3x²-75=0
<=> 3x²=75
<=> x²=25
<=> x=5
2x²-98=0
<=> 2x²=98
<=>   x²=49
<=>   x=7
x²-7x=0
<=> x(x-7)=0
<=> x=0 hoặc x=7
-3x²+5x=0
x(-3x+5)=0
x=0 hoặc -3x+5=0
x=0 hoặc -3x=-5
x=0 hoặc    x=5/3
x²+4x+4=0
(x+2)²=0
x+2=0
x=-2
 

1. 3x² - 75 = 0

<=> 3x² = 75

<=> x² = 25

<=> x = \(\sqrt{25}\)

<=> x = 5

2. x² - 7x = 0

<=> x(x - 7) = 0

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

3. x² - 14x + 13 = 0

<=> x² - 13x - x + 13 = 0

<=> x(x - 13) - (x - 13) = 0

<=> (x - 1)(x - 13) = 0

<=> \(\left[{}\begin{matrix}x-1=0\\x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=13\end{matrix}\right.\)

4. 2x² - 98 = 0

<=> 2x² = 98

<=> x² = 49

<=> x = \(\sqrt{49}\)

<=> x = 7

5. -3x² + 5x = 0

<=> x(-3x + 5) = 0

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

6. x² - 2x - 80 = 0

<=> x² + 8x - 10x - 80 = 0

<=> x(x + 8) - 10(x + 8) = 0

<=> (x - 10)(x + 8) = 0

<=> \(\left[{}\begin{matrix}x-10=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)

7. x² = 81

<=> x² - 9² = 0

<=> (x - 9)(x + 9) = 0

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

8. x² + 4x + 4 = 0

<=> x² + 2.x.2 + 2² = 0

<=> (x + 2)² = 0

<=> 0 = 0² - (x + 2)²

<=> (0 + x + 2)(0 - x + 2) = 0

<=> (x + 2)(-x + 2) = 0

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

9. 4x² + 12x + 5 = 0

<=> 4x² + 2x + 10x + 5 = 0

<=> 2x(2x + 1) + 5(2x + 1) = 0

<=> (2x + 5)(2x + 1) = 0

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

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He goes to school every morning

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His music teacher is good at singing and dancing Vietnamese traditional songs.

He watches cartoon movies on TV after doing his homework in the evening.

In evening,he watch cartoon movies on TV after doing his homework