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 \(2\cdot\widehat{C}=160^0\)

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

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

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

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

chụp rõ hơn một xíu đc ko bạn

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

\(\Leftrightarrow x^3-3x^2+27x-27-x^3+27+9x^2+18x+9=15\)

\(\Leftrightarrow45x=6\)

hay \(x=\dfrac{2}{15}\)

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

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

\(\Leftrightarrow-25x=11\)

hay \(x=-\dfrac{11}{25}\)

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

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

\(\Leftrightarrow25x=200\)

hay x=8

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

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+8+6x^2-24=60\)

\(\Leftrightarrow12x=84\)

hay x=7

6: Ta có: \(\left(x+2\right)^3-\left(x-2\right)^3=64\)

\(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8=64\)

\(\Leftrightarrow12x^2=48\)

\(\Leftrightarrow x^2=4\)

hay \(x\in\left\{2;-2\right\}\)

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

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

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

hay x=1

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

\(\Leftrightarrow16x^2+8x+1-4x^2-12x-9+5x^2+20x+20+3x^2-12=500\)

\(\Leftrightarrow20x^2+16x-500=0\)

\(\text{Δ}=16^2-4\cdot20\cdot\left(-500\right)=40256\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-16-8\sqrt{629}}{40}=\dfrac{-2-\sqrt{629}}{5}\\x_2=\dfrac{-16+8\sqrt{629}}{40}=\dfrac{-2+\sqrt{629}}{5}\end{matrix}\right.\)

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

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

\(\Leftrightarrow4x=28\)

hay x=7

Bài 3: 

1: \(35^2=1225\)

2: \(25^2=625\)

3: \(75^2=5625\)

4: \(95^2=9025\)

5: \(101\cdot99=9999\)

6: \(36\cdot44=1584\)

7: \(72\cdot68=4896\)

b: T=(-1)+(-1)+...+(-1)

=-1011

a)x=-2

b) y=10

c)x=-5

d) x=-12

\(a,\dfrac{x}{3}=\dfrac{6}{-9}\\ \Rightarrow x=-\dfrac{2}{3}.3\\ \Rightarrow x=-2\\ b,\dfrac{4}{y}=\dfrac{-2}{-5}\\ \Rightarrow y=4:\dfrac{2}{5}\\ \Rightarrow y=10\\ c,\dfrac{-2}{3}=\dfrac{x-1}{6}\\ \Rightarrow3x-3=-12\\ \Rightarrow3x=-9\\ \Rightarrow x=-3\\ d,\dfrac{3}{x}=\dfrac{6}{-24}\\ \Rightarrow x=3:-\dfrac{1}{4}\\ \Rightarrow x=-12\)

a)x=-2

b) y=10

c)x=-5

d) x=-12 .