Câu 2: 

\(2^{24}=8^8< 9^8=3^{16}\)

Câu 2: 

c: =>x+5=-4

=>x=-9

d: =>2x-3=3 hoặc 2x-3=-3

=>2x=6 hoặc 2x=0

=>x=0 hoặc x=3

Bạn ghi ra nhiều vậy người khác nhìn rối mắt không trả lời được đâu ghi từng bài ra thôi

Mình chỉ làm được vài bài thôi, kiến thức có hạn :>

Bài 1:

Câu a và c đúng

Bài 2: 

a) |x| = 2,5

=>x = 2,5 hoặc 

    x = -2,5

b) |x| = 0,56

=>x = 0,56

    x = - 0,56

c) |x| = 0

=. x = 0

d)t/tự

e) |x - 1| = 5

=>x - 1 = 5

    x - 1 = -5

f) |x - 1,5| = 2

=>x - 1,5 = 2

    x - 1,5 = -2

=>x = 2 + 1,5

    x = -2 + 1,5

=>x = 3,5

    x = - 0,5

các câu sau cx t/tự thôi

Bài 3: Ko hỉu :)

Bài 4: Kiến thức có hạn :)

\(\left|x-1,5\right|=2\\ \Rightarrow\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)

Vậy \(x\in\left\{3,5;-0,5\right\}\)

-----

\(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\\ \Rightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\\ \Rightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=-\frac{5}{4}\end{matrix}\right.\)

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

-----

\(\left|x-2\right|=x\left(ĐK:x\ge0\right)\\ \Rightarrow\left[{}\begin{matrix}x-2=x\\x-2=-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-x=2\\x+x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}0=2\left(\text{vô lý}\right)\\2x=2\end{matrix}\right.\\ \Rightarrow x=1\left(tmđk\right)\)

Vậy \(x=1\)

-----

\(\left|x-3,4\right|+\left|2,6-x\right|=0\\ \Rightarrow\left|x-3,4\right|=-\left|2,6-x\right|\)

Mà \(\left|2,6-x\right|\ge0\forall x\Rightarrow-\left|2,6-x\right|\le0\forall x\)

\(\Rightarrow\left|x-3,4\right|\le0\forall x\left(\text{vô lý}\right)\)

Vậy \(x\in\varnothing\)

a/ \(\left|x-1,5\right|=2\)

\(\Rightarrow\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2+1,5=3,5\\x=-2+1,5=-0,5\end{matrix}\right.\)

b/ \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)

\(\Rightarrow\left|x+\frac{3}{4}\right|=0+\frac{1}{2}=\frac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}-\frac{3}{4}=\frac{2}{4}-\frac{3}{4}=-\frac{1}{4}\\x=-\frac{1}{2}-\frac{3}{4}=\left(-\frac{2}{4}\right)+\left(-\frac{3}{4}\right)=-\frac{5}{4}\end{matrix}\right.\)

c/ \(\left|x-2\right|=x\)

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

\(\Rightarrow\left[{}\begin{matrix}x-x=2\\x+x=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}0=2\left(vô-lý\right)\\2x=2\end{matrix}\right.\)

=> 2x = 2

=> x = 2 : 2 = 1

d/ \(\left|x-3,4\right|+\left|2,6-x\right|=0\)

Ta có: \(\left\{{}\begin{matrix}\left|x-3,4\right|\ge0\\\left|2,6-x\right|\ge0\end{matrix}\right.\)

=> Để \(\left|x-3,4\right|+\left|2,6-x\right|=0\) thì \(\left\{{}\begin{matrix}\left|x-3,4\right|=0\\\left|2,6-x\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x-3,4=0\\2,6-x=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0+3,4=3,4\\x=2,6-0=2,6\end{matrix}\right.\)

a ) ĐK : \(x\ge0\)

Ta có : \(\left|x-2\right|=x-2\) hoặc \(\left|x-2\right|=2-x\)

TH1 : \(x-2=0\Rightarrow x=2\left(TM\right)\)

TH2 : \(2-x=0\Rightarrow x=2\left(TM\right)\)

Vậy \(x=2\)

b ) Vì \(\left|x-3,4\right|\ge0;\left|2,6-x\right|\ge0\)

\(\Rightarrow\left|x-3,4\right|+\left|2,6-x\right|\ge0\)

Để \(\left|x-3,4\right|+\left|2,6-x\right|=0\) khi \(\left|x-3,4\right|=0;\left|2,6-x\right|=0\)

\(\Rightarrow x=3,4;x=2,6\) \(\Rightarrow x=\varphi\)

a. Câu a có thể x=1 nữa.

b, \(\hept{\begin{cases}x=3,6\\x=2,6\end{cases}}\)

Nguyễn Huy Tú

dấu trị tuyệt đối phải hk bạn

a) Để (x + 1)(x - 2) < 0 thì ta có 2 trường hợp

Th1 : (x + 1) < 0 ; (x - 2) > 0 => x < -1 ; x > 2 (vô lí)

Th2 : (x + 1) > 0 ; (x - 2) < 0 => x > -1 ; x < 2 => -1 < x < 2

Vậy x thuộc {0;1}

b) Để \(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\)  thì sảy ra 2 trường hợp

Th1 : (x - 2) > 0 ; \(\left(x+\frac{2}{3}\right)>0\) => x > 2 ; \(x>-\frac{2}{3}\) => x > 2

Th2 :  (x - 2) < 0 ; \(\left(x+\frac{2}{3}\right)< 0\) => x < 2 ; \(x< -\frac{2}{3}\) => \(x< -\frac{2}{3}\)

Vậy ...........................

a,  (x+1)(x-2)<0

th1 (x+1)>0                       x>-1

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

=>  -1<x<2

TH2

      (x+1)<0

      (x-2)>0

ko xảy ra vì với mọi x nếu x-2>0=>x+1>0

Giải:

\(x-5\sqrt{x}\) = 0 (\(x\) ≥ 0)

\(\sqrt{x}\) .(\(\sqrt{x}\) - 5) = 0

\(\left[\begin{array}{l}\sqrt{x}=0\\ \sqrt{x}-5=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ \sqrt{x}=5\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=25\end{array}\right.\)

Vậy \(x\in\) {0; 25}

\(x^5\) = 2\(x^7\)

\(x^5\) - 2\(x^7\) = 0

\(x^5\).(1 - 2\(x^2\)) = 0

\(\left[\begin{array}{l}x^5=0\\ 1-2x^2=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ 2x^2=1\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x^2=\frac12\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=\pm\sqrt{\frac12}\end{array}\right.\)

Vậy \(x\) ∈ {- \(\sqrt{\frac12}\); 0; \(\sqrt{\frac12}\)}

Giải:

\(x-5\sqrt{x}\) = 0 (\(x\) ≥ 0)

\(\sqrt{x}\) .(\(\sqrt{x}\) - 5) = 0

\(\left[\begin{array}{l}\sqrt{x}=0\\ \sqrt{x}-5=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ \sqrt{x}=5\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=25\end{array}\right.\)

Vậy \(x\in\) {0; 25}

\(x^5\) = 2\(x^7\)

\(x^5\) - 2\(x^7\) = 0

\(x^5\).(1 - 2\(x^2\)) = 0

\(\left[\begin{array}{l}x^5=0\\ 1-2x^2=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ 2x^2=1\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x^2=\frac12\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=-\frac{1}{\sqrt2}\\ x=\frac{1}{\sqrt2}\end{array}\right.\)

Vậy \(x\) \(\in\) {- \(\frac{1}{\sqrt2}\); 0; \(\frac{1}{\sqrt2}\)}