Ta có:

\(g\left(x\right)=mx^2+2mx-3=0\)

Vì \(x=2\) là nghiệm của \(g\left(x\right)\) (gt).

+ Thay \(x=2\) vào \(g\left(x\right)\) ta được:

\(g\left(x\right)=m.2^2+2m.2-3=0\)

\(\Rightarrow m.4+4m-3=0\)

\(\Rightarrow4m+4m-3=0\)

\(\Rightarrow8m-3=0\)

\(\Rightarrow8m=0+3\)

\(\Rightarrow8m=3\)

\(\Rightarrow m=3:8\)

\(\Rightarrow m=\frac{3}{8}.\)

Vậy \(m=\frac{3}{8}.\)

\(g\left(x\right)=mx^2+2mx-3=0\)

Thay x = 2 vào biểu thức ta có : \(m.2^2+2m.2-3=0\Leftrightarrow4m+4m-3=0\Leftrightarrow8m-3=0\)

\(\Leftrightarrow8m=3\Leftrightarrow m=\frac{3}{8}\)

Ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{2a}{2b}=\frac{3c}{3d}=\frac{2a-3c}{2b-3d}\left(\text{đpcm}\right)\)

\(\frac{a}{b}=\frac{2a}{2b}\) 

\(\frac{c}{d}=\frac{-3c}{-3d}\) 

Áp dụng tính chất dãy tỉ số bằng nhau : 

\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}=\frac{2a-3c}{2b-3d}\)

a) |12,5 - x| = 1,3 

=> \(\orbr{\begin{cases}12,5-x=1,3\\12,5-x=-1,3\end{cases}\Rightarrow\orbr{\begin{cases}x=11,2\\x=13,8\end{cases}}}\)

b) \(\left|x-\frac{2}{3}\right|+\frac{5}{4}=0\Rightarrow\left|x-\frac{2}{3}\right|=-\frac{5}{4}\)

Vì \(\left|x-\frac{2}{3}\right|\ge0\text{ mà }-\frac{5}{4}< 0\Rightarrow x\in\varnothing\)

a, \(\left|12,5-x\right|=1,3\Leftrightarrow\orbr{\begin{cases}12,5-x=1,3\\12,5-x=-1,3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=11,2\\x=13,8\end{cases}}\)

b, \(\left|x-\frac{2}{3}\right|+\frac{5}{4}=0\Leftrightarrow \left|x-\frac{2}{3}\right|=-\frac{5}{4}\)( vô lí )

\(\left|x-\frac{2}{3}\right|\ge0;-\frac{5}{4}< 0\)

a) \(\sqrt{0,01}+\sqrt{0,16}-12^0=0,1+0,4-1=-0,5\)

b) \(\left(\frac{1}{16}\right)^7:\left(\frac{1}{8}\right)^9=\left[\left(\frac{1}{2}\right)^4\right]^7:\left[\left(\frac{1}{2}\right)^3\right]^9=\left(\frac{1}{2}\right)^{28}:\left(\frac{1}{2}\right)^{27}=\frac{1}{2}\)

a) 

\(=\sqrt{0,1^2}+\sqrt{0,4^2}-1\) 

\(=|0,1|+|0,4|-1\) 

\(=0,1+0,4-1=-0,5\) 

b) 

= \(\left(\left(\frac{1}{2}\right)^4\right)^7:\left(\left(\frac{1}{2}\right)^3\right)^9\) 

=\(\left(\frac{1}{2}\right)^{4\cdot7}:\left(\frac{1}{2}\right)^{3\cdot9}\) 

= \(\left(\frac{1}{2}\right)^{28}:\left(\frac{1}{2}\right)^{27}\) 

\(=\left(\frac{1}{2}\right)^{28-27}\)     

\(=\left(\frac{1}{2}\right)^1=\frac{1}{2}\)       

a) xⁿ.x^m = x^{n + m}

b) \(\frac{x^n}{x^m}=x^n:x^m=x^{n-m}\)

c) \(\left(x^n\right)^m=x^{n.m}\)

d) \(\left(x.y\right)^n=x^n,y^n\)

e) \(\left(\frac{x}{y}\right)^n=\frac{x^n}{y^n}\)

\(\frac{3}{4}\left(\frac{2}{5}\right)^{14}:\left(\frac{4}{25}\right)^6=\frac{3}{4}\left(\frac{2}{5}\right)^{14}:\left(\frac{2}{5}\right)^{2.6}=\frac{3}{4}\left(\frac{2}{5}\right)^2=\frac{3}{4}.\frac{4}{25}=\frac{3}{25}\)

a) \(x^n.x^m=x^{n+m}\)

b) \(\frac{x^n}{x^m}=x^n\div x^m=x^{n-m}\)

c) \(\left(x^n\right)^m=x^{n.m}\)

d) \(\left(x.y\right)^n=x^n.y^n\)

e) \(\left(\frac{x}{y}\right)^n=\frac{x^n}{y^n}\left(y\ne0\right)\)

Áp dụng \(\frac{3}{4}.\left(\frac{2}{5}\right)^{14}\div\left(\frac{4}{25}\right)^6=\frac{3}{4}.\frac{2^{14}}{5^{14}}\div\frac{4^6}{25^6}\)

                                                          \(=\frac{3}{4}.\frac{2^{14}}{5^{14}}.\frac{25^6}{4^6}\)

                                                          \(=\frac{3.2^{14}.\left(5^2\right)^6}{4.5^{14}.\left(2^2\right)^6}=\frac{3.2^{14}.5^{12}}{2^2.5^{14}.2^{12}}=\frac{3}{25}\)

a) Xét tứ giác BECD có : 

M là trung điểm ED 

M là trung điểm BC

=》 BECD là hình bình hành 

=》BE//DC 

b) Vì BECD là hình bình hành 

=》EC//BD 

Mà NBD = 90°

Lại có : NBD + CNB = 180°

=》 CNB = 90°

Vậy CN\(\perp\)AB 

Hay CE\(\perp\)AB

Bạn làm lại theo cách làm của hình Tam Giác giúp mình

a) Ta có: \(A=\left|3x+\frac{1}{3}\right|-\frac{1}{4}\ge-\frac{1}{4}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left|3x+\frac{1}{3}\right|=0\Rightarrow x=-\frac{1}{9}\)

Vậy Min(A) = -1/4 khi x = -1/4

b) Ta có: \(\frac{3}{4}-\left|2x-\frac{1}{2}\right|0\le\frac{3}{4}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left|2x-\frac{1}{2}\right|=0\Rightarrow x=\frac{1}{4}\)

Vậy Max(B) = 3/4 khi x = 1/4

a. Vì \(\left|3x+\frac{1}{3}\right|\ge0\forall x\)\(\Rightarrow A=\left|3x+\frac{1}{3}\right|-\frac{1}{4}\ge-\frac{1}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\left|3x+\frac{1}{3}\right|=0\Leftrightarrow3x+\frac{1}{3}=0\Leftrightarrow x=-\frac{1}{9}\)

Vậy minA = - 1/4 <=> x = - 1/9

b. Vì \(\left|2x-\frac{1}{2}\right|\ge0\forall x\)\(\Rightarrow B=\frac{3}{4}-\left|2x-\frac{1}{2}\right|\le\frac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\left|2x-\frac{1}{2}\right|=0\Leftrightarrow2x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}\)

Vậy maxB = 3/4 <=> x = 1/4

Bài làm:

Ta có: \(a=15^{120}\div25^{60}\)

\(a=15^{120}\div5^{120}\)

\(a=3^{120}=9^{60}\)

và \(b=2^{45}.2^{15}.4^{60}\)

\(b=2^{60}.2^{120}\)

\(b=2^{180}=8^{60}\)

Mà \(9^{60}>8^{60}\Rightarrow a>b\)

    \(\frac{1}{2.15}+\frac{3}{2.11}+\frac{4}{1.11}+\frac{5}{1.2}\)

\(=\frac{1}{30}+\left(\frac{3}{22}+\frac{4}{11}\right)+\frac{5}{2}\)

\(=\frac{1}{30}+\frac{1}{2}+\frac{5}{2}\)

\(=\frac{1}{30}+3\)

\(=\frac{91}{30}\)

= 3 \(\frac{1}{30}\)Hoặc  =\(\frac{91}{30}\)