a) \(\left|x\right|=2,5\Rightarrow\)\(\left[{}\begin{matrix}x=2,5\\x=-2,5\end{matrix}\right.\)

b) \(\left|x\right|=-1,2\left(VLý\right)\Rightarrow S=\varnothing\)

c) \(\left|x\right|+0,573=2\Rightarrow\left|x\right|=1,427\)

\(\Rightarrow\left[{}\begin{matrix}x=1,427\\x=-1,427\end{matrix}\right.\)

d) \(\left|x+\dfrac{1}{3}\right|-4=-1\Rightarrow\left|x+\dfrac{1}{3}\right|=3\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{100}{3}\end{matrix}\right.\)

mở dấu trị tuyệt đối ra rồi tính như bình thường

a) vì /x/>= 0 => x = { 2,5 ; -2,5 }

b) ko tìm đươc x thỏa mãn vì /x/ >= 0

c) /x/ = 2 + 0,573

<=> /x/ = 2,573

<=> x = { 2,573 ; -2,573 }

d) /x+ 1/3 / = -1+(-4 )

<=> /x+1/3 /= -5

vì /x+1/3 / luôn lớn hơn hoặc bằng 0 => ko tìm được x thỏa mãn

a) \(2^x=8\)

⇔ \(2^x=2^3\)

⇒ \(x=3\)

b) \(3^x=27\)

⇔ \(3^x=3^3\)

⇒ \(x=3\)

c) \(\left(-\dfrac{1}{2}\right)x=\left(-\dfrac{1}{2}\right)^4\)

⇔ \(x=\left(-\dfrac{1}{2}\right)^4\div\left(-\dfrac{1}{2}\right)\)

⇔ \(x=\left(-\dfrac{1}{2}\right)^3\)

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

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

⇔ \(x=\left(-\dfrac{3}{4}\right)^3=-\dfrac{27}{64}\)

d) \(\left(x+1\right)^3=-125\)

⇔ \(\left(x+1\right)^3=\left(-5\right)^3\)

⇔ \(x+1=-5\)

⇔ \(x=-5-1=-6\)

2:

a: (x-1,2)^2=4

=>x-1,2=2 hoặc x-1,2=-2

=>x=3,2(loại) hoặc x=-0,8(loại)

b: (x-1,5)^2=9

=>x-1,5=3 hoặc x-1,5=-3

=>x=-1,5(loại) hoặc x=4,5(loại)

c: (x-2)^3=64

=>(x-2)^3=4^3

=>x-2=4

=>x=6(nhận)

a) |x| = 2,5

=>\(\left[\begin{array}{nghiempt}x=2,5\\x=-2,5\end{array}\right.\)

vậy x=2,5 hoặc x=-2,5

b)|x|=-1,2

=>x không có giá trị thỏa mãn |x|\(\ge\) 0

c)|x| + 0,573 = 2

|x| = 2 - 0,573

|x| = 1,427

=>\(\left[\begin{array}{nghiempt}x=1,427\\x=-1,427\end{array}\right.\)

Vậy x = 1,427 hoặc x = -1,427

d) ∣∣x+13∣∣ - 4 = -1

=>|x+\(\frac{1}{3}\)| =-1 + 4

|x+\(\frac{1}{3}\)| = 3

.....................

Vậy x = \(\frac{8}{3}\) hoặc x = \(\frac{-10}{3}\)

a ) \(\left|x\right|=2,5\Rightarrow x=2,5;x=-2,5\)

b ) \(\left|x\right|=-1,2\Rightarrow\left|x\right|\ge0\forall x\Rightarrow x\in\varnothing\)

c ) \(\left|x\right|+0,573=2\)

\(\Rightarrow\)\(\left[\begin{array}{nghiempt}x+0,573=2\\x+0,573=-2\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}2-0,573\\\left(-2\right)-0,573\end{array}\right.\) \(\Rightarrow\)\(\left[\begin{array}{nghiempt}x=1,427\\x=-2,573\end{array}\right.\)

Vậy \(x\in1,427;-2,573\)

d ) \(\left|x+\frac{1}{3}\right|-4=-1\)

\(\Rightarrow\left|x+\frac{1}{3}\right|=3\)

\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{3}=3\\x+\frac{1}{3}=-3\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=3-\frac{1}{3}\\x=\left(-3\right)-\frac{1}{3}\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{8}{3}\\x=\frac{-10}{3}\end{array}\right.\)

Vậy \(x\in\frac{8}{3};\frac{-10}{3}\)

(Gíup mình với các thần đồng 's hoc24 ơi)

\(a,\left[0,75x+\frac{5}{2}\right]-\frac{4}{7}-\left[-\frac{1}{3}\right]=-\frac{5}{6}\)

\(\Leftrightarrow\left[0,75x+\frac{5}{2}\right]-\frac{4}{7}+\frac{1}{3}=-\frac{5}{6}\)

\(\Leftrightarrow\left[0,75x+\frac{5}{2}\right]=-\frac{5}{6}-\frac{1}{3}+\frac{4}{7}\)

\(\Leftrightarrow0,75x+\frac{5}{2}=-\frac{25}{42}\)

\(\Leftrightarrow0,75x=-\frac{65}{21}\Leftrightarrow x=-\frac{260}{63}\)

\(b,\left[4x-9\right]\left[2,5+-\frac{7}{3}x\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x-9=0\\2,5+-\frac{7}{3}x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=\frac{15}{14}\end{cases}}\)

Vậy \(x\in\left\{\frac{9}{4};\frac{15}{14}\right\}\)

Tham khảo nhé phamthiminhanh

a) \(\left(0,75x+\frac{5}{2}\right)-\frac{4}{7}-\left(-\frac{1}{3}\right)=-\frac{5}{6}\)

=>   \(0,75x+\frac{5}{2}-\frac{4}{7}+\frac{1}{3}=-\frac{5}{6}\)

=> \(0,75x=-\frac{5}{6}-\frac{5}{2}+\frac{4}{7}-\frac{1}{3}\)

=>  \(0,75x=-\frac{65}{21}\)

=>   \(x=-\frac{260}{63}\)

b) \(\left(4x-9\right)\left(2,5+-\frac{7}{3}x\right)=0\)

=>   \(10x-\frac{28}{3}x^2-22,5+21x=0\)

=>   \(-\frac{28}{3}x^2+31x-22,5=0\)

=>   Phần còn lại bạn tự giải tiếp nhé!

a: \(x^3-4x^2-x+4=0\)

=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)

=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

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

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)

=>\(\left(x+3\right)\left(x^2-4\right)=0\)

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

=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)

=>\(\left(x-2\right)\left(x-2-4\right)=0\)

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)