1C

2A

1C        2A

a) \(\left(x^2-3\right)\left(x^2-36\right)=0\)

TH1: \(x^2-3=0\Rightarrow x^2=3\)

Ta thấy không có số nguyên nào mà bình phương nên bằng 3 nên không có giá trị x thỏa mãn.

TH2: \(x^2-36=0\Rightarrow x^2=36=6.6=\left(-6\right).\left(-6\right)\)

Vậy x = 6 hoặc x = -6.

b) \(\left(x^2-3\right)\left(x^2-36\right)< 0\)

Do \(x^2-3>x^2-36\) nên chỉ có thể xảy ra trường hợp \(\hept{\begin{cases}x^2-3>0\\x^2-36< 0\end{cases}}\)

\(\Rightarrow3\le x^2\le36\Rightarrow2\le x\le6\) hoặc \(-6\le x\le-2\)

1a) x=6;x=-6;x=-cang3;x=cang3

1b)x<6

3(x + 2)^2 + (2x - 1)^2 - 7(x + 3)(x - 3) = 36

=> 3(x^2 + 4x + 4) + 4x^2 - 4x + 1 - 7(x^2 - 9) = 36

=> 3x^2 + 12x + 12 + 4x^2 - 4x + 1 - 7x^2 + 63 = 36

=> 8x + 76 = 36

=> 8x = -40

=> x = -5

Bạn ghi đề lại câu a.

3(x−1)^2−3x(x−5)=1

⇒3(x^2−2x+1)−3x^2+15x=1

⇒3x^2−6x+3−3x^2+15x=1

=9x+3=1

⇒9x=(−3)+1

⇒x=−2/9 

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

\(\Rightarrow3\left(x^2-2x+1\right)-3x^2+15x=1\)

\(\Rightarrow3x^2-6x+3-3x^2+15x=1\)

\(=9x+3=1\)

\(\Rightarrow9x=\left(-3\right)+1\)

\(\Rightarrow x=\frac{-2}{9}\)

Cảm ơn nhiều nhá

\(a)x=\dfrac{1}{4}+\dfrac{5}{13}=\dfrac{33}{52}.\\ b)\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}.\\ \Leftrightarrow\dfrac{x}{3}=\dfrac{11}{21}.\\ \Leftrightarrow\dfrac{7x}{21}=\dfrac{11}{21}.\\ \Rightarrow7x=11.\\ \Leftrightarrow x=\dfrac{11}{7}.\\ c)\dfrac{x}{3}=\dfrac{16}{24}+\dfrac{24}{36}=\dfrac{2}{3}+\dfrac{2}{3}=\dfrac{4}{3}.\\ \Rightarrow x=4.\\ d)\dfrac{x}{15}=\dfrac{1}{5}+\dfrac{2}{3}=\dfrac{13}{15}.\\ \Rightarrow x=13.\)

b)(x+3)²-(x-4)(x+8)=1

\(\Rightarrow\)x²+6x+9-(x²+8x-4x-32)=1

⇒x²+6x+9-x²-8x+4x+32=1

⇒2x+41=1

\(\Rightarrow\)2x+41-1=0

\(\Rightarrow\)2x+40=0

⇒2x=-40

\(\Rightarrow\)x=\(\dfrac{-40}{2}\)

⇒x=-20

\(a,\left(2x+1\right)^2-4\left(x+2\right)^2=9\\ \Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)-9=0\\ \Leftrightarrow4x^2-4x^2+4x-16x+1-16-9=0\\ \Leftrightarrow-12x=24\\ \Leftrightarrow x=\dfrac{24}{-12}=-2\\ b,\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\\ \Leftrightarrow x^2+6x+9-\left(x^2+4x-32\right)=1\\ \Leftrightarrow x^2-x^2+6x-4x=1-9-32\\ \Leftrightarrow2x=-40\\ \Leftrightarrow x=-20\\ c,3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\\ \Leftrightarrow3\left(x^2+4x+4\right)+\left(4x^2-4x+1\right)-7\left(x^2-9\right)=36\\ \Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2+63=36\\ \Leftrightarrow3x^2+4x^2-7x^2+12x-4x=36-12-1-63\\ \Leftrightarrow8x=-40\\ \Leftrightarrow x=\dfrac{-40}{8}=-5\)