\(1024:2^6+160:\left(3^3+53\right)-9^{100}:9^{99}\)

\(=2^{10}:2^6+160:\left(26+53\right)-9\)

\(=2^4+160:80-9\)

\(=16+2-9\)

\(=9\)

saii đề bài ak,

Tìm x:1. / x+1/+/3-2x/=/3x-2/

2. /x-3/-2/5-2x/=11

Ko có giá trị tuyệt đối nên mk dùng tạm dấu gạch ngang

nhầm ak,tìm x: |x+3|=|5−2x|

\(\left|x+3\right|=\left|5-2x\right|\)

TH1 : \(x+3=5-2x\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)

TH2 : \(x+3=2x-5\Leftrightarrow x=8\)

4tvq333333t43q

\(a,\frac{x+2}{-15}=\left(-\frac{5}{3}\right)\)

\(\Leftrightarrow\left(-\frac{x+2}{15}\right)=\left(-\frac{5}{3}\right)\)

\(\Leftrightarrow\left(-3\right)\times\left[-\left(x+2\right)\right]=15\times5\)

\(\Leftrightarrow3x+6=75\)

\(\Leftrightarrow3x=69\)

\(\Leftrightarrow x=23\)

\(a)\)

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

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

\(\Leftrightarrow f\left(x\right)=-x^3+2x^2+x-3+8\)

\(\Leftrightarrow f\left(x\right)=-x^3+2x^2+x+5\)

\(g\left(x\right)=-3x^3-x^2+2x^3+5x-3-4x\)

\(\Leftrightarrow g\left(x\right)=\left(-3x^3+2x^3\right)-x^2+\left(5x-4x\right)-3\)

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

\(b)\)

Theo đề ra: \(h\left(x\right)=f\left(x\right)+g\left(x\right)\)

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

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

\(\Rightarrow h\left(x\right)=x^2+2x+2\)

tính góc gì thế bạn

ACD^ ah

a, \(\left|x-1\right|=3x+2\)ĐK : x >= -2/3 

TH1 : \(x-1=3x+2\Leftrightarrow2x=-3\Leftrightarrow x=-\frac{3}{2}\)( ktm )

TH2 : \(x-1=-3x-2\Leftrightarrow4x=-1\Leftrightarrow x=-\frac{1}{4}\)( tm )

b, \(\left|5x\right|=x-12\)ĐK : x > = 12 

TH1 : \(5x=x-12\Leftrightarrow4x=-12\Leftrightarrow x=-3\)( ktm )

TH2 : \(5x=12-x\Leftrightarrow6x=12\Leftrightarrow x=2\)( ktm )

c, \(\left|7-x\right|=5x+1\)ĐK : x >= -1/5

TH1 : \(7-x=5x+1\Leftrightarrow6x=-6\Leftrightarrow x=-1\)( ktm )

TH2 : \(7-x=5x+1\Leftrightarrow6x=6\Leftrightarrow x=1\)( tm )

a) Điều kiện : \(x\ge-\frac{2}{3}\)

Xét : 

• \(-\frac{2}{3}\le x< 1\)

\(\Rightarrow x-1< 0\)

\(\Rightarrow\left|x-1\right|=1-x\)

\(\Rightarrow1-x=3x+2\)

\(\Rightarrow4x=-1\)

\(\Rightarrow x=-\frac{1}{4}\left(TM\right)\)

• \(x\ge1\)

\(\Rightarrow x-1\ge0\)

\(\Rightarrow\left|x-1\right|=x-1\)

\(\Rightarrow x-1=3x+2\)

\(\Rightarrow2x=-3\)

\(\Rightarrow x=-\frac{3}{2}\left(KTM\right)\)

Vậy \(x=-\frac{1}{4}\)

b) Điều kiện : \(x\ge12\)

\(\Rightarrow5x>0\)

\(\Rightarrow\left|5x\right|=5x\)

\(\Rightarrow5x=x-12\)

\(\Rightarrow4x=-12\)

\(\Rightarrow x=-3\left(KTM\right)\)

Vậy không có x

c)Điều kiện : \(x\ge-\frac{1}{5}\)

Xét :

• \(-\frac{1}{5}\le x\le7\)

\(\Rightarrow7-x\ge0\)

\(\Rightarrow\left|7-x\right|=7-x\)

\(\Rightarrow7-x=5x+1\)

\(\Rightarrow6x=6\)

\(\Rightarrow x=1\left(TM\right)\)

• \(x>7\)

\(\Rightarrow7-x< 0\)

\(\Rightarrow\left|7-x\right|=x-7\)

\(\Rightarrow x-7=5x+1\)

\(\Rightarrow4x=-8\)

\(\Rightarrow x=-2\left(KTM\right)\)

Vậy \(x=1\)