1C

2A

Câu 1: A

Câu 2: C

l) (x + 9) . (x² – 25) = 0  

<=> (x + 9) . (x – 5) . (x + 5) = 0   

<=> \(\left[{}\begin{matrix}\text{x + 9 = 0}\\x-5=0\\x+5=0\end{matrix}\right.\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)

Vậy S = \(\left\{-9,5,-5\right\}\)

      e) |x - 4 |< 7         

<=> \(\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)

Vậy S = \(\left\{11;-3\right\}\)

I,(x+9).(x^2-25)=0

tương đương:x+9=0

                       x^2-25=0

tương đương : x=-9

                       x=5

e,\(\left|x-4\right|\)=7

tương đương x-4=4

                       x-4=-4

tương đương :x=0

                        x=-8

Bài 1: 

l) Ta có: \(\left(x+9\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{-9;5;-5\right\}\)

e) Ta có: |x-4|<7

mà \(\left|x-4\right|\ge0\forall x\)

nên \(\left|x-4\right|\in\left\{0;1;2;3;4;5;6\right\}\)

\(\Leftrightarrow x-4\in\left\{0;1;-1;2;-2;3;-3;4;-4;5;-5;6;-6\right\}\)

hay \(x\in\left\{4;5;3;6;2;7;1;8;0;9;-1;10;-2\right\}\)

Vậy: \(x\in\left\{4;5;3;6;2;7;1;8;0;9;-1;10;-2\right\}\)

f) Ta có: \(40< 31+\left|x\right|< 47\)

\(\Leftrightarrow\left|x\right|+31\in\left\{41;42;43;44;45;46\right\}\)

\(\Leftrightarrow\left|x\right|\in\left\{10;11;12;13;14;15\right\}\)

hay \(x\in\left\{10;-10;11;-11;12;-12;13;-13;-14;14;15;-15\right\}\)

Vậy: \(x\in\left\{10;-10;11;-11;12;-12;13;-13;-14;14;15;-15\right\}\)

g) Ta có: \(\left|x+3\right|\le2\)

\(\Leftrightarrow\left|x+3\right|\in\left\{0;1;2\right\}\)

\(\Leftrightarrow x+3\in\left\{0;1;-1;2;-2\right\}\)

hay \(x\in\left\{-3;-2;-4;-1;-5\right\}\)

Vậy: \(x\in\left\{-3;-2;-4;-1;-5\right\}\)

Lời giải:

a. 

$(25-2x)^3:5-3^2=4^2$

$(25-2x)^3:5=4^2+3^2=25$

$(25-2x)^3=25.5=5^3$

$\Rightarrow 25-2x=5$

$\Rightarrow 2x=20$

$\Rightarrow x=10$

b.

$2.3^x=10.3^{12}+8.27^4=10.3^{12}+8.3^{12}=18.3^{12}=2.3^{14}$

$\Rightarrow 3^x=3^{14}$

$\Rightarrow x=14$

\(11\times\left(x-6\right)=4\times x+11\)

\(\Rightarrow11\times x-66=4\times x+11\)

\(\Rightarrow11\times x-4\times x=66+11\)

\(\Rightarrow77\times x=77\)

\(\Rightarrow x=1\)

b) \(\left(2\times x-10\right)\times\left(25-25\times x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2\times x-10=0\\25-25\times x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}}\)

a)11x(X-6)=4xX+11

   11xX-11x6=4xX+11

    11xX-4xX=11x6+11

    (11-4)xX=66+11

      7xX=77

       x=77:7

       x=11

b)(2xX-10)x(25-25xX)=0

=>2xX-10=0=>2xX=10 =>x=10:2=5

hoặc 25-25xX=0=>25xX=25=>x=25:25=1

Vậy x=1 hoặc x=5

Tích ch0 mình nha

Cảm ơn bạn nhìu

a)11x(X-6)=4xX+11

   11xX-11x6=4xX+11

    11xX-4xX=11x6+11

    (11-4)xX=66+11

      7xX=77

       x=77:7

       x=11

b)(2xX-10)x(25-25xX)=0

=>2xX-10=0=>2xX=10 =>x=10:2=5

hoặc 25-25xX=0=>25xX=25=>x=25:25=1

Vậy x=1 hoặc x=5

a)(x/10 -3/2)^2 =1/25 

TH1 : x/10 -3/2 =1/5

=> x/10 =17

=> x=170

TH2 :x/10 -3/2 =-1/5

x/10 =1,3

x=13

Vậy x =170 hoặc 13

B) |x-1|+4=15

=>|x-1|=11

TH1:x-1=11=> x=12

TH2: x-1=-11 => x=-10

Vậy x=12 hoặc -10

Đáp số:

a) x=45

b) x=31

c) x=2

d) x=94

a) x = 15

b) x = 11

c) x = 68

d) x = 310

(x4)3=x6x19

a) x = 15.              b)      x = 11.        c) x = 68.              d) x = 310

Tìm xx biết:    \left(x^{4}\right)^{3}=\dfrac{x^{19}}{x^{6}}(x4)3=x6x19​

Trả lời: x=x=