?

1.C

2.B

3.C

4.A

8.A

9.B

10.C

1/ \(2\left(x-5\right)=\left(-x-5\right)\)

\(\Leftrightarrow2x-10=-x-5\)

\(\Leftrightarrow3x=5\)

\(\Leftrightarrow x=\dfrac{5}{3}\)

Vậy: \(S=\left\{\dfrac{5}{3}\right\}\)

==========

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

\(\Leftrightarrow2x+6-3x+3=2\)

\(\Leftrightarrow-x=-7\)

\(\Leftrightarrow x=7\)

Vậy: \(S=\left\{7\right\}\)

==========

3/ \(4\left(x-5\right)-\left(3x-1\right)=x-19\)

\(\Leftrightarrow4x-20-3x+1=x-19\)

\(\Leftrightarrow0x=0\)

Vậy: \(S=\left\{x|x\text{ ∈ }R\right\}\) 

===========

4/ \(7-\left(x-2\right)=5\left(2-3x\right)\)

\(\Leftrightarrow7-x+2=10-15x\)

\(\Leftrightarrow14x=1\)

\(\Leftrightarrow x=\dfrac{1}{14}\)

Vậy: \(S=\left\{\dfrac{1}{14}\right\}\)

==========

5/ \(2x-\left(5-3x\right)=7x+1\)

\(\Leftrightarrow2x-5+3x=7x+1\)

\(\Leftrightarrow-2x=6\)

\(\Leftrightarrow x=-3\)

Vậy: \(S=\left\{-3\right\}\)

[---]

Chúc bạn học tốt.

a/ \(\left(x+y\right)^2-8\left(x+y\right)+12\)

\(=\left(x+y\right)\left(x+y-8+12\right)\)

\(=\left(x+y\right)\left(x+y+4\right)\)

==========

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

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

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

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

===========

c/ \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)

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

\(=\left(x^2+x\right)\left(x^2+x-17\right)\)

[---]

27.A

28.B

29.B

30.D

31.B

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△ABC vuông cân tại A ⇒ \(AB=AC=4\left(cm\right)\)

Xét △ABC vuông tại A

\(BC^2=AB^2+AC^2\left(Pytago\right)\)

\(\Rightarrow BC=\sqrt{4^2+4^2}=\sqrt{36}\left(cm\right)\)

Do △ACD vuông cân tại D ⇒ AC là cạnh huyền, \(AD=CD\)

- Xét △ACD vuông tại D:

\(AC^2=AD^2+CD^2\left(Pytago\right)\)

- Mà \(AD=CD\)

\(\Rightarrow4^2=AD^2+CD^2=2^2+2^2\)

\(\Rightarrow AD=2\left(cm\right);CD=2\left(cm\right)\)

- Mặt khác: \(\hat{ADC}=90\text{°}\left(gt\right)\)

\(\hat{BCD}=\hat{ACB}+\hat{ACD}=45\text{°}+45\text{°}=90\text{°}\)

⇒ AD//BC ⇒ABCD là hình thang

⇒ ABCD là hình thang vuông, có đường cao CD

\(S_{ABCD}=\dfrac{\left(4+\sqrt{32}\right).2}{2}=4+\sqrt{32}\approx9,7\left(cm^2\right)\)

Vậy: \(S_{ABCD}\approx9,7cm^2\)

[----------]

\(16x^2-\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(4x+x+1\right)\left(4x-x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+x+1=0\Leftrightarrow5x=-1\Leftrightarrow x=-\dfrac{1}{5}\\4x-x-1=0\Leftrightarrow3x=1\Leftrightarrow x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{5};\dfrac{1}{3}\right\}\)

1/ \(7x-5=13-5x\)

\(\Leftrightarrow12x=18\)

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

Vậy: \(S=\left\{\dfrac{3}{2}\right\}\)

==========

2/ \(19+3x=5-18x\)

\(\Leftrightarrow21x=-14\)

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

Vậy: \(S=\left\{-\dfrac{2}{3}\right\}\)

==========

3/ \(x^2+2x-4=-12+3x+x^2\)

\(\Leftrightarrow-x=-8\)

\(\Leftrightarrow x=8\)

Vậy: \(S=\left\{8\right\}\)

===========

4/ \(-\left(x+5\right)=3\left(x-5\right)\)

\(\Leftrightarrow-x-5=3x-15\)

\(\Leftrightarrow-4x=-10\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

Vậy: \(S=\left\{\dfrac{5}{2}\right\}\)

==========

5/ \(3\left(x+4\right)=\left(-x+4\right)\)

\(\Leftrightarrow3x+12=-x+4\)

\(\Leftrightarrow4x=-8\)

\(\Leftrightarrow x=-2\)

Vậy: \(S=\left\{-2\right\}\)

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