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.

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

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

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

`A=-x^2+2x+10`

`=-(x^2-2x)+10`

`=-(x-1)^2+11<=11`

Dấu "=" xảy ra khi `x=1`.

`B=4x-2x^2+8`

`=-2(x^2-2x)+8`

`=-2(x^2-2x+1)+10`

`=-2(x-1)^2+10<=10`

Dấu "=" xảy ra khi `x=1`

`C=-x^2-x+1`

`=-(x^2+x)+1`

`=-(x^2+x+1/4)+1+1/4`

`=-(x+1/2)^2+5/4<=5/4`

Dấu "=" xảy ra khi `x=-1/2`

`D=-4x^2+6x+3`

`=-(4x^2-6x)+3`

`=-(4x^2-6x+9/4)+21/4`

`=-(2x-3/2)^2+21/4<=21/4`

Dấu "=' xảy ra khi `2x=3/2<=>x=3/4`

\(a,A=-x^2+2x+10=-x^2+2x-1+11=-\left(x^2-2x+1\right)+11\)

\(=11-\left(x-1\right)^2\)

- Thấy : \(\left(x-1\right)^2\ge0\forall x\in R\)

\(\Rightarrow A=11-\left(x-1\right)^2\le11\)

Vậy MaxA = 11 <=> x = 1 .

\(b,B=-2x^2+4x-2+10=-2\left(x^2-2x+1\right)+10=10-2\left(x-1\right)^2\)

- Thấy : \(\left(x-1\right)^2\ge0\forall x\in R\)

\(\Rightarrow B=10-2\left(x-1\right)^2\le10\)

Vậy MaxB = 10 <=> x = 1 .

\(c,C=-x^2-\dfrac{1}{2}.2.x-\dfrac{1}{4}+\dfrac{5}{4}=\dfrac{5}{4}-\left(x+\dfrac{1}{2}\right)^2\)

- Thấy : \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\in R\)

\(\Rightarrow C=\dfrac{5}{4}-\left(x+\dfrac{1}{2}\right)^2\le\dfrac{5}{4}\)

Vậy MaxC = 5/4 <=> x = -1/2 .

\(d,D=-4x^2+6x+3=-4x^2+2x.2.\dfrac{6}{4}-\dfrac{9}{4}+\dfrac{21}{4}=-\left(4x^2-6x+\dfrac{9}{4}\right)+\dfrac{21}{4}\)

\(=\dfrac{21}{4}-\left(2x-\dfrac{3}{2}\right)^2\)

- Thấy : \(\left(2x-\dfrac{3}{2}\right)^2\ge0\forall x\in R\)

\(\Rightarrow A=\dfrac{21}{4}-\left(2x-\dfrac{3}{2}\right)^2\le\dfrac{21}{4}\)

Vậy MaxD=21/4 <=> x = 3/4 .

a: =>x^2-25-x^2-3x=10

=>-3x=35

=>x=-35/3

b: =>4x^2-9-4(x^2+4x+4)=5

=>4x^2-9-4x^2-16x-16-5=0

=>-16x-30=0

=>x=-15/8

c: =>9x^2+45x-9x^2+4=7

=>45x=3

=>x=1/15

d: =>x^3+3x^2+3x+1-x^3-3x^2+5x=8

=>8x=7

=>x=7/8

1: 7-x=8+(-7)

=>7-x=8-7=1

=>x=7-1=6

2: \(x-8=\left(-3\right)-8\)

=>x-8=-11

=>\(x=-11+8=-3\)

3: \(2-x=10-9+23\)

=>\(2-x=33-9=24\)

=>x=2-24=-22

4: \(-2-x=15\)

=>\(x=-2-15=-17\)

5: \(-7+x-8=-3-1+13\)

=>x-14=13-4=9

=>x=9+14=23

6: 100-x+7=-x+3

=>107-x=3-x

=>107=3(vô lý)

7: \(23+x=8-2x\)

=>\(x+2x=8-23\)

=>3x=-15

=>x=-15/3=-5

1) \(ĐK:x\in R\)

2) \(ĐK:x< 0\)

3) \(ĐK:x\in\varnothing\)

4) \(=\sqrt{\left(x+1\right)^2+2}\) 

\(ĐK:x\in R\)

5) \(=\sqrt{-\left(a-4\right)^2}\)

\(ĐK:x\in\varnothing\)

Tìm x biết :a) ( 2x - 3 ).( x +1 ) > 0b) ( x + 5 ).(x-7) < 0c) | 2x - 3 | + 8 = 10d) ( 2x + 5 ) . | x -8 | . ( x2 + 1 ) = 0

\(\dfrac{5}{x}+1+\dfrac{4}{x}+1=\dfrac{3}{-13}\\ \Rightarrow\dfrac{9}{x}+2=-\dfrac{3}{13}\\ \Rightarrow\dfrac{9}{x}=-\dfrac{59}{13}\\ \Rightarrow x=-\dfrac{207}{59}\)

a. \(\dfrac{5}{x+1}+\dfrac{4}{x+1}=\dfrac{-3}{13}\)

ĐKXĐ: x ≠ -1

⇔ \(\dfrac{65}{13\left(x+1\right)}+\dfrac{52}{13\left(x+1\right)}=\dfrac{-3\left(x+1\right)}{13\left(x+1\right)}\)

⇔ 65 + 52 = -3(x + 1)

⇔ 117 = -3x - 3

⇔ 117 + 3 = -3x

⇔ 120 = -3x 

⇔ x = \(\dfrac{120}{-3}=-40\) (TM)

b. -x + 2 + 2x + 3 + x + \(\dfrac{1}{4}\) + 2x + \(\dfrac{1}{6}\) = \(\dfrac{8}{3}\)

⇔ -x + 2x + x + 2x = \(\dfrac{8}{3}-\dfrac{1}{6}-\dfrac{1}{4}-3-2\)

⇔ 4x = -2,75

⇔ x = \(\dfrac{-2,75}{4}=\dfrac{-11}{16}\)

c. \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+2}\) = \(\dfrac{12}{26}\)

⇔  \(\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{2\left(3x+1\right)}=\dfrac{12}{26}\)

⇔ \(\dfrac{312\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) + \(\dfrac{520\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) - \(\dfrac{312\left(2x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)

= \(\dfrac{48\left(2x+1\right)\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)

⇔ 312(3x +1) + 520(3x + 1) - 312(2x + 1) = 48(2x + 1)(3x + 1)

⇔ 936x + 312 + 1560x + 520 - 624x - 312 = (96x + 48)(3x + 1)

⇔ 936x + 312 + 1560x + 520 - 624x - 312 = 288x² + 96x + 144x + 48

⇔ 936x + 1560x - 624x - 96x - 144x - 288x² = 48 - 312 - 520 + 312

⇔ 1632x - 288x² = -472

⇔ -288x² + 1632x + 472 = 0 (Tự giải tiếp, dùng phương pháp tách hạng tử)

⇔ x = 5,942459684 \(\approx\) 6

c: Ta có: \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)

\(\Leftrightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Leftrightarrow2x+1=13\)

hay x=6

1: =>3^x=81

=>x=4

2: =>2^x=8

=>x=3

3: =>x^3=2^3

=>x=2

4: =>x^20-x=0

=>x(x^19-1)=0

=>x=0 hoặc x=1

5: =>2^x=32

=>x=5

6: =>(2x+1)^3=9^3

=>2x+1=9

=>2x=8

=>x=4

7: =>x^3=115

=>\(x=\sqrt[3]{115}\)

8: =>(2x-15)^5-(2x-15)^3=0

=>(2x-15)^3*[(2x-15)^2-1]=0

=>2x-15=0 hoặc (2x-15)^2-1=0

=>2x-15=0 hoặc 2x-15=1 hoặc 2x-15=-1

=>x=15/2 hoặc x=8 hoặc x=7

1. Tìm số tự nhiên x biết:

1) \(3^x.3=243\)

\(3^x=243:3\)

\(3^x=81\)

\(3^x=3^4\)

\(\Rightarrow x=4\)

_____

2) \(7.2^x=56\)

\(2^x=56:7\)

\(2^x=8\)

\(2^x=2^3\)

\(\Rightarrow x=3\)

_____

3) \(x^3=8\)

\(x^3=2^3\)

\(\Rightarrow x=3\)

_____

4) \(x^{20}=x\)

\(x^{20}-x=0\)

\(x\left(x^{19}-1\right)=0\)

\(\Rightarrow x=0\) hoặc \(x=1\)

5) \(2^x-15=17\)

\(2^x=17+15\)

\(2^x=32\)

\(2^x=2^5\)

\(\Rightarrow x=5\)

_____

6) \(\left(2x+1\right)^3=9.81\)

\(\left(2x+1\right)^3=729=9^3\)

\(\rightarrow2x+1=9\)

\(2x=9-1\)

\(2x=8\)

\(x=8:2\)

\(\Rightarrow x=4\)

_____

7) \(x^6:x^3=125\)

\(x^3=125\)

\(x^3=5^3\)

\(\Rightarrow x=5\)

_____

8) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=7\\x=8\end{matrix}\right.\)

_____

9) \(3^{x+2}-5.3^x=36\)

\(3^x.\left(3^2-5\right)=36\)

\(3^x.\left(9-5\right)=36\)

\(3^x.4=36\)

\(3^x=36:4\)

\(3^x=9\)

\(3^x=3^2\)

\(\Rightarrow x=2\)

_____

10) \(7.4^{x-1}+4^{x+1}=23\)

\(\rightarrow7.4^{x-1}+4^{x-1}.4^2=23\)

\(4^{x-1}.\left(7+4^2\right)=23\)

\(4^{x-1}.\left(7+16\right)=23\)

\(4^{x-1}.23=23\)

\(4^{x-1}=23:23\)

\(4^{x-1}=1\)

\(4^{x-1}=4^1\)

\(\rightarrow x-1=0\)

\(x=0+1\)

\(\Rightarrow x=1\)

Chúc bạn học tốt

\(\left|5x+13\right|=2x-7\)

khi \(x>\frac{7}{2}\), biểu thức có dạng:

\(\orbr{\begin{cases}5x+13=2x-7\\5x+13=7-2x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=-20\\7x=-6\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{20}{3}\\x=-\frac{6}{7}\end{cases}}}\)

ta có 

a. (5x-7)(x-9)-(-x+3)(-5x+2)= 2x(x-4)-(x-1)(2x+3)

\(\Leftrightarrow5x^2-52x+63-\left(5x^2-17x+6\right)=2x^2-8x-\left(2x^2+x-3\right)\)

\(\Leftrightarrow-35x+57=-9x+3\Leftrightarrow26x=54\Leftrightarrow x=\frac{27}{13}\)

b. (x-3)(-x+10)+(x-8)(x+3)= (5x^2-1)(x+3)-5x^3-15x^2

\(\Leftrightarrow-x^2+13x-30+x^2-5x-24=5x^3+15x^2-x-3-5x^3-15x^2\)

\(\Leftrightarrow8x-54=-x-3\Leftrightarrow9x=51\Leftrightarrow x=\frac{17}{3}\)