b,x=10

a, x=2029

Xin lỗi, mình sai đề :

a) (x² - 1) . ( x² - 4) < 0

\(\text{b) }2+4+6+...+2x=210\\ \Leftrightarrow2\left(1+2+3+...+x\right)=210\\ \Leftrightarrow1+2+3+...+x=105\\ \Leftrightarrow1+2+3+...+x=1+2+3+...+14\\\Leftrightarrow\left(1+2+3+...\right)+x=\left(1+2+3+...\right)+14\\ \Leftrightarrow x=14 \)

Vậy \(x=14\)

5: (x-1)(x-5)=(x-1)(x-2)

=>(x-1)(x-5)-(x-1)(x-2)=0

=>(x-1)(x-5-x+2)=0

=>-3(x-1)=0

=>x-1=0

=>x=1

6: \(6\left(x-3\right)\left(x-4\right)-6x\left(x-2\right)=4\)

=>\(6\left(x^2-7x+12\right)-6x^2+12x=4\)

=>\(6x^2-42x+72-6x^2+12x=4\)

=>-30x=4-72=-68

=>\(x=\frac{68}{30}=\frac{34}{15}\)

7: -(x+3)(x-4)+(x+1)(x-1)=10

=>\(-\left(x^2-x-12\right)+x^2-1=10\)

=>\(-x^2+x+12+x^2-1=10\)

=>x+11=10

=>x=-1

8: (2x-1)(x-2)-(2x-7)(x+3)=3

=>\(2x^2-4x-x+2-\left(2x^2+6x-7x-21\right)=3\)

=>\(2x^2-5x+2-\left(2x^2-x-21\right)=3\)

=>\(2x^2-5x+2-2x^2+x+21=3\)

=>-4x+23=3

=>-4x=-20

=>x=5

9: \(\left(x-5\right)\left(4-x\right)-\left(x-1\right)\left(x+3\right)=-2x^2\)

=>\(-\left(x^2-9x+20\right)-\left(x^2+2x-3\right)=-2x^2\)

=>\(-x^2+9x-20-x^2-2x+3=-2x^2\)

=>7x-17=0

=>7x=17

=>x=17/7

10: (4x+1)(x-3)-(x-7)(4x-1)=15

=>\(4x^2-12x+x-3-\left(4x^2-x-28x+7\right)=15\)

=>\(4x^2-11x-3-4x^2+29x-7=15\)

=>18x-10=15

=>18x=25

=>x=25/18

11: \(\left(x+1\right)\left(x^2-x+1\right)-x\left(x^2-3\right)=4\)

=>\(x^3+1-x^3+3x=4\)

=>3x=4-1=3

=>x=1

12: \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)=6x\)

=>\(x^3-27+5x-x^3=6x\)

=>6x=5x-27

=>x=-27

13: (3x-5)(x+1)-(3x-1)(x+1)=x-4

=>(x+1)(3x-5-3x+1)=x-4

=>-4(x+1)=x-4

=>-4x-4=x-4

=>-4x-x=0

=>-5x=0

=>x=0

14; 5(x-3)(x-7)-(5x+1)(x-2)=8

=>\(5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=8\)

=>\(5x^2-50x+105-5x^2+9x+2=8\)

=>-41x=8-2-105=6-105=-99

=>x=99/41

a) \(\left|x-1\right|+\left|x+3\right|=4\left(1\right)\)

+) TH1: Nếu \(x< -3\) thì \(x-1< 0;x+3< 0\)

\(\Rightarrow\left|x-1\right|=-x+1;\left|x+3\right|=-x-3\)

PT (1) trở thành: \(-x+1-x-3=4\)

\(\Leftrightarrow-2x=6\Leftrightarrow x=-3\left(loại\right)\)

+) TH2: Nếu \(-3\le x< 1\) thì \(x-1< 0;x+3>0\)

\(\Rightarrow\left|x-1\right|=-x+1;\left|x+3\right|=x+3\)

PT (1) trở thành: \(-x+1+x+3=4\)

\(\Leftrightarrow0x=0\) (luôn đúng)

Kết hợp với đk ta được: \(\Rightarrow-3\le x< 1\)

+) TH3: Nếu \(x\ge1\) thì \(x-1>0;x+3>0\)

\(\Rightarrow\left|x-1\right|=x-1;\left|x+3\right|=x+3\)

PT (1) trở thành: \(x-1+x+3=4\)

\(\Leftrightarrow2x=2\Leftrightarrow x=1\left(t/m\right)\)

Vậy x nằm trong khoảng \(-3\le x\le1.\)

Mấy bài kia làm tương tự.

2.

\(\left|x+1\right|+\left|x+2\right|+...+\left|x+10\right|=605x\)(1)

Vì các thừa số ở vế phải của (1) đều không âm nên x không âm. Do đó \(\left|x+1\right|+\left|x+2\right|+...+\left|x+10\right|=\left(x+1\right)+\left(x+2\right)+...+\left(x+10\right)\)

\(\Rightarrow\left(x+1\right)+\left(x+2\right)+...+\left(x+10\right)=605x\)

\(\Rightarrow10x+\dfrac{10\left(10+1\right)}{2}=605x\)

\(\Rightarrow55=595x\)

\(\Rightarrow x=\dfrac{55}{595}=\dfrac{11}{119}\)

Vậy x = \(\dfrac{11}{119}\)

Hình như ở câu b còn thiếu trường hợp bằng -1 thì phải?

=> \(\left(2x-15\right)^3\left(2x-15-1\right)\left(2x-15+1\right)=0\)

=> \(\left(2x-15\right)^3\left(2x-16\right)\left(2x-14\right)=0\)

=> \(\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{15}{2}\\x=8\\x=7\end{matrix}\right.\)

Vậy ...

a) \(\left(2x-1\right)^3=-8\)

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

\(\Leftrightarrow2x-1=2\)

\(\Leftrightarrow2x=2+1\)

\(\Leftrightarrow2x=3\)

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

Tick mình nha

1,

\(\left(2x+1\right)^3=-0,001\\ \left(2x+1\right)^3=\left(-0.1\right)^3\\ \Leftrightarrow2x+1=-0.1\\ 2x=-1.1\\ x=-\dfrac{11}{10}:2\\ x=-\dfrac{11}{20}\\ Vậy...\)

2,

\(\left(2x-3\right)^4=\left(2x-3\right)^6\\ \Leftrightarrow\left(2x-3\right)^6-\left(2x-3\right)^4=0\\ \Leftrightarrow\left(2x-3\right)^4\cdot\left[\left(2x-3\right)^2-1\right]=0\\ \Rightarrow\left\{{}\begin{matrix}\left(2x-3\right)^4=0\\\left(2x-3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\\left(2x-3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x=3\\2x-3=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\\ Vậyx\in\left\{\dfrac{3}{2};2\right\}\)

3, Làm tương tự câu 2

5,

\(9^x:3^x=3\\ \left(9:3\right)^x=3\\ 3^x=3\\ \Rightarrow x=1\\ Vậy...\)

6,

\(3^x+3^{x+3}=756\\ 3^x+3^x\cdot3^3\\ 3^x\cdot\left(1+27\right)=756\\ 3^x\cdot28=756\\ \Leftrightarrow3^x=27\\ 3^x=3^3\\ \Rightarrow x=3\\ vậy...\)

7,

\(5^{x+1}+6\cdot5^{x+1}=875\\ 5^{x+1}\cdot\left(1+6\right)=875\\ 5^{x+1}\cdot7=875\\ \Leftrightarrow5^{x+1}=125\\ \Leftrightarrow5^{x+1}=5^3\Leftrightarrow x+1=3\\ \Rightarrow x=2\\ Vậy...\)

9,

lê thị hồng vân trả lời típ đi