=5 nka bn k mk nka 

5 

kết bn mik nhé

\(\left(2x^2+1\right)\left(4x-3\right)=\left(2x^2+1\right)\left(x-13\right)\)

\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3-x+13\right)=0\)

\(\Leftrightarrow\left(2x^2+1\right)\left(3x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2+1=0\\3x+10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=-\dfrac{1}{2}\left(VN\right)\\x=-\dfrac{10}{3}\end{matrix}\right.\)

\(S=\left\{-\dfrac{10}{3}\right\}\)

\(\left(2x^2+1\right)\left(4x-3\right)=\left(2x^2+1\right)\left(x-12\right)\)

\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3\right)-\left(2x^2+1\right)\left(x-12\right)=0\)
\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3-x+12\right)=0\)

\(\Leftrightarrow\left(2x^2+1\right)\left(3x+9\right)=0\)

\(\Leftrightarrow3x+9=0\) (do \(2x^2+1>0\forall x\in R\))

\(\Leftrightarrow x=-3\)

-Vậy \(S=\left\{-3\right\}\)

Câu 1:

a: 3(x-1)-(x+1)=-1

=>3x-3-x-1=-1

=>2x-4=-1

=>2x=-1+4=3

=>\(x=\frac32\)

b: Đặt f(x)=0

=>\(2x^2-x=0\)

=>x(2x-1)=0

=>x=0 hoặc x=1/2

Bài 3:

a: Xét ΔCDA và ΔEAD có

CD=EA

\(\hat{CDA}=\hat{EAD}\) (hai góc so le trong, CD//AE)

AD chung

Do đó: ΔCDA=ΔEAD

b: Ta có: \(\hat{BAD}+\hat{CAD}=\hat{BAC}=90^0\)

\(\hat{BDA}+\hat{HAD}=90^0\) (ΔHAD vuông tại H)

mà \(\hat{CAD}=\hat{HAD}\) (AD là phân giác của góc HAC)

nên \(\hat{BAD}=\hat{BDA}\)

=>ΔBAD cân tại B

Bài 2:

a: g(x)-f(x)+h(x)

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

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

b: f(x)=2x^2-3x-x+1=2x^2-4x+1

f(-1)=\(2\cdot\left(-1\right)^2-4\cdot\left(-1\right)+1=2+4+1=7\)

\(h\left(\frac12\right)=2\cdot\left(\frac12\right)^2+1=2\cdot\frac14+1=\frac12+1=\frac32\)

f(-1)-h(1/2)

=7-3/2

=11/2

c: f(x)=h(x)

=>\(2x^2-4x+1=2x^2+1\)

=>-4x=0

=>x=0

Chọn D

H(-1) = 2\(x^2\)- 10 

H(-1) = 2.(-1)² - 10

H(-1) = 2 - 10

H(-1) = -8

H(\(\dfrac{1}{2}\)) = 2\(x^2\) - 10

H(\(\dfrac{1}{2}\)) = 2.(\(\dfrac{1}{2}\))² - 10

H(\(\dfrac{1}{2}\)) = \(\dfrac{1}{2}\) - 10

H(\(\dfrac{1}{2}\)) = \(-\dfrac{19}{2}\)

A   = − 6 x 5 + 4 x 4 + 2 x 3 − 2 x 2 + 2 x 3 − 6 x 2 + 2 x

A   =   − 6 x 5 + 4 x 4 + 4 x 3 − 8 x 2 + 2 x

Chọn đáp án C

a) \(M=\frac{2\times2}{1\times5}+\frac{2\times2}{5\times9}+\frac{2\times2}{9\times13}+...+\frac{2\times2}{45\times40}\)

\(M=\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{45\times49}\)

\(M=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{45}-\frac{1}{49}\)

\(M=1-\frac{1}{49}\)

\(M=\frac{48}{49}\)

b) \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+5+...+10}\)

=  \(\frac{2}{2\times\left(1+2\right)}+\frac{2}{2\times\left(1+2+3\right)}+...+\frac{2}{2\times\left(1+2+3+...+10\right)}\)

\(=\frac{2}{6}+\frac{2}{12}+...+\frac{2}{110}\)

\(=\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{10\times11}\)

\(=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(=2\times\left(\frac{1}{2}-\frac{1}{11}\right)\)

\(=2\times\frac{9}{22}\)

\(=\frac{9}{11}\)

Mình trả lời câu a nha  M= 4/1*5+4/5*9+4/9*13+...+4/45*49   M=1-1/5+1/5-1/9+1/9-1/13+...+1/45-1/49    M=1-1/49=48/49