\(-x^4+4x^2-5x^2+20=0\\\Rightarrow -(x^4-4x^2)-(5x^2-20)=0\\\Rightarrow-x^2(x^2-4)-5(x^2-4)=0\\\Rightarrow(x^2-4)(-x^2-5)=0\\\Rightarrow-(x-2)(x+2)(x^2+5)=0\\\Rightarrow(2-x)(x+2)=0(vì.x^2+5>0\forall x)\)

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

4(x - 10) - (-3x + 4) = -2(x + 5) - (-4x +20)

=> 4x - 40 + 3x - 4 = -2x - 10 + 4x - 20

=> 7x - 44 = 2x - 30

=> 7x - 2x = -30 + 44

=> 5x = 14

=> x = 14 : 5

=> x = 14/5

\(4\left(x-10\right)-\left(-3x+4\right)=-2\left(x+5\right)-\left(-4x+20\right)\)

\(\Rightarrow4x-40+3x-4=-2x-10+4x-20\)

\(\Rightarrow4x+3x+2x-4x=-10-20+4+40\)

\(\Rightarrow5x=14\Rightarrow x=\frac{14}{5}\)

\(4(x-10)-(-3x+4)=-2(x+5)-(-4x+20)\)

\(\Rightarrow4x-40+3x-4=(-2)x-10+4x-20\)

\(\Rightarrow7x-44=2x-30\)

\(\Rightarrow7x-2x=44-30\)

\(\Rightarrow5x=14\)

\(\Rightarrow x=\frac{14}{5}\)

\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)

\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)

Bài 1:

a. $(-20)+x=-30$

$x-20=-30$

$x=-30+20=-(30-20)=-10$

b.

$(-10)-x=-20$

$x=(-10)-(-20)=-10+20=20-10=10$

c. Đề sai. Bạn xem lại.

d.

$x+(-3)=-7$

$x=-7-(-3)=-7+3=-(7-3)=-4$

e.

$x-(-5)=-9$

$x=(-9)+(-5)=-14$

f.

$x(-11)=12$

$x=\frac{12}{-11}=\frac{-12}{11}$

h.

$2x-10=20$

$2x=20+10=30$

$x=30:2=15$

l.

$4x-8=-8$
$4x=-8+8=0$

$x=0:4=0$

k.

$-12-(-2)x=-8$

$(-2)x=-12-(-8)=-12+8=-(12-8)=-4$

$x=(-4):(-2)=2$

Bài 2:

a. $-20-(10-x)=-3$

$10-x=-20-(-3)=-20+3=-(20-3)=-17$

$x=10-(-17)=10+17=27$

b.

$14+(14-x)=-2$
$14-x=-2-14=-16$

$x=14-(-16)=14+16=30$

c.

$-15-(x-3)=-7$

$x-3=-15-(-7)=-15+7=-8$

x=-8+3=-5$

d.

$(x+4)+(-20)=-8$

$x+4=-8-(-20)=-8+20=12$
$x=12-4=8$

e.

$-2x-2=-4$

$-2x=-4+2=-2$

$x=(-2):(-2)=1$

f.

$-2x+4=-4$

$-2x=-4-4=-8$

$x=(-8):(-2)=4$

l.

$-12-(-2)x=-2-4=-6$

$(-2)x=-12-(-6)=-12+6=-6$

$x=(-6):(-2)=3$

5x-16=40+x

=> 5x-16-x = 40

=> 5x-x -16=40

4x-16=40

4x= 40+16

4x=56

x= 56:4

x=14

Vậy...

4x-10=15-x

=> 4x-10+x= 15

4x+x -10=15

5x= 15+10

5x= 25

x= 25:5

x=5

Vậy....

5x -16=40+x

=> 5x-x=40+16

=>4x=56

=>x=56:4

x=14

Câu 6: Khôg có cau nào đúng

Câu 7: C

Câu 8: B

Câu 9: B

Câu 10: D

6Không có đáp án nào đúng x=11/4

7C

8B

9B

10D

\(\dfrac{4}{3}\) = \(\dfrac{20}{4}\)\(x\) - 1

5\(x\) - 1 = \(\dfrac{4}{3}\)

5\(x\)      = \(\dfrac{4}{3}\) + 1

5\(x\)     = \(\dfrac{7}{3}\)

 \(x\)      = \(\dfrac{7}{3}\) : 5

 \(x\)     = \(\dfrac{7}{15}\)

b,

2\(\times\)\(x\) = 3\(\times\) y 

     \(x\) = \(\dfrac{3}{2}\) \(\times\) y

y - \(\dfrac{3}{2}\) \(\times\) y = 5

\(y\) \(\times\) ( 1 - \(\dfrac{3}{2}\)) = 5

\(y\) \(\times\) - \(\dfrac{1}{2}\) = 5

\(y\)            = 5 : (-\(\dfrac{1}{2}\))

\(y\)            = - 10

\(x\)  = y - 5 = -10 - 5 =-15