a, => x= -9. 2

=> x= -18

b, => x= 12. (-4)

=> x=-48

c, => 152-(3x-1)=54

=> (3x-1)= 152- 54= 98

=> 3x= 98+1= 99

=> x = 99:3

=> x = 33

tick cho mik nha

Ta có:

a) \(x:2=-9\Rightarrow x=\left(-9\right).2=-18\)

b)\(x:\left(-4\right)=12\Rightarrow x=12.\left(-4\right)=-48\)

c)\(x+x+80=-8\Rightarrow2x+80=-8\Rightarrow2x=-8-80=-88\Rightarrow x=\left(-88\right):2=-44\)d)\(152-\left(3x-1\right)=\left(-2\right).\left(-27\right)=54\Rightarrow152-54=3x-1\Rightarrow98=3x-1\Rightarrow3x=99\Rightarrow x=33\)

Ta có \(A\left(1\right)=a+b+c+d=A\left(-1\right)=a+b-c+d\)

=> c = -c (1)

Ta có \(A\left(2\right)=16+4b+2c+d=A\left(-2\right)=16a+4b-2c+d\)

=> 2c = -2c (2)

Từ (1) và (2) => A(x) = A(-x)

a) -105 - 5.x = (-5)^2

=>-105 - 5.x = 25

=> 5.x = -105 - 25

=> 5.x = -130

=> x = -130: 5

=> x = -26

c) 400 -4.|5-x| = (-6)^2

=>400- 4.|5-x| = 36

=> 4.|5-x| = 400-36

=> 4.|5-x| = 364

=> |5-x| = 364:4

=> |5-x| = 91

=> \(\left[\begin{matrix}5-x=91\\5-x=-91\end{matrix}\right.\)

=> \(\left[\begin{matrix}x=5-91\\x=5-\left(-91\right)\end{matrix}\right.\)

^{=> \(\left[\begin{matrix}x=-86\\x=96\end{matrix}\right.\)}

d) 2.|x-1| -5 = -15

=>2.|x-1| = -15+5

=>2.|x-1| = 10

=> |x-1| = 10:2

=> |x-1| = 5

=> \(\left[\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\)

=>

Bài 1.

a, (x-2)-15=65

x-2=65+15

x-2=80

x=80+2

x=2

b, 115-2\(\times\)(x-3)=35

2\(\times\)(x-3)=115-35

2\(\times\)(x-3)=70

x-3=70:2

x-3=35

x=35+5

x=38

c, 35+2\(\times\)(x-3)=65

2\(\times\)(x-3)=65-35

2\(\times\)(x-3)=30

x-3=30:2

x-3=15

x=15+3

x=18

3\(\times\)(x-5)-16=11

3\(\times\)(x-5)=11+16

3\(\times\)(x-5)=27

x-5=27:3

x-5=9

x=9+5

x=14

Bài 2:

a, \(2^x-1=31\)

\(2^x=31-1\)

\(2^x=30\)

\(\Rightarrow\)Không có x thoả mãn điều kiện \(2^x=30\)

b, \(x^3-1=26\)

\(x^3=26+1\)

\(x^3=27\)

\(\Rightarrow x=3\) vì \(3^3=27\)

c, \(6^x-1+1=37\)

\(6^x-1=37-1\)

\(6^x-1=36\)

\(6^x=36+1\)

\(6^x=37\)

\(\Rightarrow\) Không có x thoả mãn điều kiện \(6^x=37\)

d, (x+2)\(^3\)-15\(^0\)=215

\(\left(x+2\right)^3-1=215\)

\(\left(x+2\right)^3=215+1\)

\(\left(x+2\right)^3=216\)

\(\left(x+2\right)^3=6^3\)

\(x+2=6\)

\(x=6-2\)

\(x=4\)

e, \(2\times\left(x-9\right)^2=2\)

\(\left(x-9\right)^2=2:2\)

\(\left(x-9\right)^2=1\)

\(\Rightarrow x-9=1\) vì \(1^2=1\)

x=1+9

x=10

g, \(3\times\left(x-5\right)^3=51\)

\(\left(x-5\right)^3=51:3\)

\(\left(x-5\right)^3=17\)

\(\Rightarrow\) Không có x thoả mãn điều kiện \(\left(x-5\right)^3=17\)

Nếu đúng thì tick cho mk nhé

Bài 1:

a)\(\left(x-2\right)-15=65\)

\(x-2=65+15\)

\(x-2=80\)

\(x=80+2\)

\(x=82\)

b)\(115-2\left(x-3\right)=35\)

\(2\left(x-3\right)=115-35\)

\(2\left(x-3\right)=80\)

\(x-3=80:2\)

\(x-3=40\)

\(x=40+3\)

c) \(35+2\left(x-3\right)=65\)

\(2\left(x-3\right)=65-35=30\)

\(x-3=30:2=15\)

\(x=15+3=18\)

d) \(3\left(x-5\right)-16=11\)

\(3\left(x-5\right)=11+16=27\)

\(x-5=27:3=9\)

\(x=9+5=14\)

Bài 2:

a) \(2^x-1=31\)

\(2^x=31+1=32\)

Vì \(2^5=32\Rightarrow x=5\)

b) \(x^3-1=26\)

\(x^3=26+1=27\)

Vì \(3^3=27\Rightarrow x=3\)

c)\(6^{x-1}+1=37\)

\(6^{x-1}=37-1=36\)

Vì \(6^6=36\Rightarrow x-1=6\Rightarrow x=6+1=7\)

d)\(\left(x+2\right)^3-15^0=215\)

\(\left(x+2\right)^3-1=215\)

\(\left(x+2\right)^3=215+1=216\)

Vì \(6^3=216\Rightarrow x+2=6\Rightarrow x=6-2=4\)

e)\(2\left(x-9\right)^2=2\)

\(\left(x-9\right)^2=2:2=1\)

Vì \(1^2=1\Rightarrow x-9=1\Rightarrow x=1+9=10\)

g) \(3\left(x-5\right)^3=51\)

\(\left(x-5\right)^3=51:3=17\)

Bài 1:

a)

\(\left(x-2\right)-15=65\)

\(x-2\) \(=65+15\)

\(x-2\) \(=\) \(80\)

\(x\) \(=80+2\)

\(x\) \(=82\)

b)

\(115-2\cdot\left(x-3\right)=35\)

\(2\cdot\left(x-3\right)=115-35\)

\(2\cdot\left(x-3\right)=80\)

\(x-3=80:2\)

\(x-3=40\)

\(x\) \(=40+3\)

\(x\) \(=43\)

c)

\(35+2\cdot\left(x-3\right)=65\)

\(2\cdot\left(x-3\right)=65-35\)

\(2\cdot\left(x-3\right)=30\)

\(x-3=30:2\)

\(x-3=15\)

\(x\) \(=15+3\)

\(x\) \(=18\)

d)

\(3\cdot\left(x-5\right)-16=11\)

\(3\cdot\left(x-5\right)\) \(=11+16\)

\(3\cdot\left(x-5\right)\) \(=27\)

\(x-5\) \(=27:3\)

\(x-5\) \(=9\)

\(x\) \(=9+5\)

\(x\) \(=14\)

Bài 2

a)

\(2^x-1=31\)

\(2^x\) \(=31+1\)

\(2^x\) \(=32\)

\(2^x\) \(=2^5\)

⇒ \(x=5\)

b)

\(x^3-1=26\)

\(x^3\) \(=26+1\)

\(x^3\) \(=27\)

\(x^3\) \(=3^3\)

⇒ \(x=3\)

c)

\(6^{x-1}+1=37\)

\(6^{x-1}\) \(=37-1\)

\(6^{x-1}\) \(=36\)

\(6^{x-1}\) \(=6^2\)

➞ \(x-1=2\)

\(x\) \(=2+1\)

\(x\) \(=3\)

d)

\(\left(x+2\right)^3-15^0=215\)

\(\left(x+2\right)^3-1=215\)

\(\left(x+2\right)^3=215+1\)

\(\left(x+2\right)^3=216\)

\(\left(x+2\right)^3=6^3\)

➞ \(x+2=6\)

\(x=6-2\)

\(x=4\)

e)

\(2\cdot\left(x-9\right)^2=2\)

\(\left(x-9\right)^2=2:2\)

\(\left(x-9\right)^2=1\)

\(\left(x-9\right)^2=1^2\)

➞ \(x-9=1\)

\(x=9+1\)

\(x=10\)

g)

\(3\cdot\left(x-5\right)^3=51\)

Câu hỏi này sai nhé bạn

Bài 2 :

a, x = \(\dfrac{-3}{-11}\) => x =\(\dfrac{3}{11}\)

=>| x | = \(\dfrac{3}{11}\)

=> x= \(\dfrac{3}{11}\) hoặc x = \(\dfrac{-3}{11}\)

Bài 3 :

a, | 4.(x-1)| =12

=> 4.(x-1)=12 hoặc 4.(x-1)=-12

\(\left[{}\begin{matrix}4.\left(x-1\right)=12\\4.\left(x-1\right)=-12\end{matrix}\right.=>\left[{}\begin{matrix}4x-4=12\\4x-4=-12\end{matrix}\right.=>\left[{}\begin{matrix}4x=16\\4x=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

Vậy x = 4 hoặc x = -2

b,|2x+1|-5 =10

|2x+1|=15

=2x+1=15 hoặc 2x+=-15

+) 2x+1=15 = > 2x = 14 = > x =7

+)2x+1=-15 => 2x= -16 => x = -8

Vậy x=7 hoặc x = -8

c,|2,5-x|-1,3=0

|2,5-x|= 1,3

=>2,5 -x = 1,3 hoặc 2,5 - x = -1,3

+)2,5 - x = 1,3 => x = 1,2

+)2,5-x = -1,3 => x=3,8

Vậy x = 1,2 hoặc x = 3,8

d,-|1,4 - x | - 2 = 0

-|1,4-x|=2

=> -1,4+x = 2 hoặc -1,4+x = -2

+) -1,4+x= 2 => x = 3,4

+)-1,4+x= -2 => x = 0,6

Vậy x = 3,4 hoặc x = 0 ,6

e,| x - 2 | = x

=> x -2 = x hoặc x - 2 = -x

+) x- 2 = x => x-x = -2 => 0 = -2 ( vô lí )

+) x -2 = -x => x+x=2 => 2x =2 => x= 1

Vậy x = 1

f, 2.|2x-3| = \(\dfrac{1}{2}\)

=> |2x-3|= \(\dfrac{1}{4}\)

=>2x-3=\(\dfrac{1}{4}\) hoặc 2x-3=\(\dfrac{-1}{4}\)

+) 2x - 3 = \(\dfrac{1}{4}\)=> 2x= \(\dfrac{13}{4}\)=> x = \(\dfrac{13}{8}\)

+) 2x - 3 = \(\dfrac{-1}{4}\)=> 2x=\(\dfrac{11}{4}\)=> x = \(\dfrac{11}{8}\)

Vậy x=\(\dfrac{13}{8}\) hoặc x=\(\dfrac{11}{8}\)

B1:a )

(x-4).(y+3)=-3=-1.3=-3.1

ta có bảng sau:

Lần sau ghi tách ra tí bạn ơi ;v

--------------------------------

1. a) \(5x-20y=5\left(x-4y\right)\)

b) \(5\left(x-1\right)-3x\left(x-1\right)=\left(x-1\right)\left(5-3x\right)\)

c) \(x\left(x+1\right)-5x-5=x\left(x+1\right)-5\left(x+1\right)\)

\(=\left(x+1\right)\left(x-5\right)\)

d) \(\left(x+y\right)^2-\left(x-y\right)^2\)

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

\(=4xy\)

e) \(\left(3x+1\right)^2-\left(x+1\right)^2\)

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

\(=2x\left(4x+2\right)\)

2. a) \(x+5x^2=0\)

\(\Leftrightarrow x\left(1+5x\right)=0\)

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

Vậy...

b) \(x+1=\left(x+1\right)^2\)

\(\Leftrightarrow x+1-x^2-2x-1=0\)

\(\Leftrightarrow-x^2-x=0\)

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

\(\Rightarrow\left[{}\begin{matrix}-x=0\\x+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Vậy...

c) \(x^3+x=0\)

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

Vì \(x^2+1>0\Rightarrow x=0\)

Vậy...

d) \(x^3-0,25x=0\)

\(\Leftrightarrow x\left(x^2-0,25\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-0,25=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm0,5\end{matrix}\right.\)

Vậy..

e) \(x^2-10x=-25\)

\(\Leftrightarrow x^2-10x+25=0\)

\(\Leftrightarrow\left(x-5\right)^2=0\)

\(\Rightarrow x-5=0\)

\(\Rightarrow x=5\)

Vậy...

a: \(\left|x+\dfrac{4}{15}\right|-\left|-3.75\right|=-\left|-2.5\right|\)

\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.5+3.75=1.25=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=\dfrac{5}{4}\\x+\dfrac{4}{15}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{59}{60}\\x=-\dfrac{91}{60}\end{matrix}\right.\)

c: \(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)

d: Ta có: \(\left|x\left(x^2-\dfrac{5}{4}\right)\right|=x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x\left(x^2-\dfrac{5}{4}\right)=x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x\left(x^2-\dfrac{9}{4}\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;\dfrac{3}{2}\right\}\)