(x+2)(4x^2 -2x+1)+(3-2x)(9+6x+4x^2) =  -4x^3+6x^2-3x+29

nha bạn chúc bạn học tốt nha 

Trả lời:

( x + 2 )( 4x² - 2x + 1 ) + ( 3 - 2x )( 9 + 6x + 4x² )

= 4x³ - 2x² + x + 8x² - 4x + 2 + 27 + 18x + 12x² - 18x - 12x² - 8x³ 

= - 4x³ + 6x² - 3x + 29

a) ĐKXĐ: x <= 2/3

Pt --> 2 - 3x = 4

<=> 3x = -2

<=> x = -2/3 (thỏa)

b) ĐKXĐ: x >= 2

Pt --> x^2 + 4x + 4 = x^2 - 4x + 4

<=> 8x = 0<=> x = 0(loại)

a: Ta có: \(\sqrt{2-3x}=2\)

\(\Leftrightarrow2-3x=4\)

\(\Leftrightarrow3x=-2\)

hay \(x=-\dfrac{2}{3}\)

b: Ta có: \(\sqrt{x^2+4x+4}=x-2\)

\(\Leftrightarrow\left|x+2\right|=x-2\)

\(\Leftrightarrow x+2=2-x\left(x< -2\right)\)

\(\Leftrightarrow x=0\left(loại\right)\)

Chứng minh gì á bạn?

a) \(3x^2+y^2-2xy+4x+20=\left(x^2-2xy+y^2\right)+2\left(x^2+2x+1\right)+18=\left(x-y\right)^2+2\left(x+1\right)^2+18\ge18>0\forall x,y\)

\(ĐTXR\Leftrightarrow x=y=-1\)

a) ĐKXĐ: x <= 2

pt --> 4 - 2x = 25 <=> x = -21/2 (thỏa)

??

Đề kiểu gì vậy ?

b) ĐKXĐ: x >= -1

pt <=> 8sqrt(x + 1)=16 <=> sqrt(x+1)=2 --> x + 1 = 4 <=> x = 3

a, ĐKXĐ: \(x\le2\)

\(\sqrt{4-2x}=5\\ \Leftrightarrow4-2x=25\\ \Leftrightarrow2x=-21\\ \Leftrightarrow x=-10,5\left(tm\right)\)

b, ĐKXĐ: \(x\ge-1\)

\(\sqrt{25\left(x+1\right)}+\sqrt{9x+9}=16\\ \Leftrightarrow5\sqrt{x+1}+\sqrt{9\left(x+1\right)}=16\\ \Leftrightarrow5\sqrt{x+1}+3\sqrt{x+1}=16\\ \Leftrightarrow8\sqrt{x+1}=16\\ \Leftrightarrow\sqrt{x+1}=2\\ \Leftrightarrow x+1=4\\ \Leftrightarrow x=3\)

c, \(\sqrt{4x^2+12x+9}=4\Leftrightarrow4x^2+12x+9=16\\ \Leftrightarrow4x^2+12x-7=0\\ \Leftrightarrow\left(4x^2-2x\right)+\left(14x-7\right)=0\\ \Leftrightarrow2x\left(2x-1\right)+7\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

a: \(\Leftrightarrow4-2x=25\)

hay \(x=-\dfrac{21}{2}\)

c: \(\Leftrightarrow\left|2x+3\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=4\\2x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

a,ĐKXĐ:\(x\ge2\)

\(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\dfrac{x-2}{4}}=26\\ \Leftrightarrow4\sqrt{x-2}+3\sqrt{x-2}-\dfrac{\sqrt{x-2}}{2}=26\\ \Leftrightarrow8\sqrt{x-2}+6\sqrt{x-2}-\sqrt{x-2}=52\\ \Leftrightarrow13\sqrt{x-2}=52\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)

b,ĐKXĐ:\(x\in R\)

\(3x+\sqrt{4x^2-8x+4}=1\\ \Leftrightarrow2\sqrt{x^2-2x+1}=1-3x\\ \Leftrightarrow\left|x-1\right|=\dfrac{1-3x}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1-3x}{2}\\x-1=\dfrac{3x-1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-2=1-3x\\2x-2=3x-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

c, ĐKXĐ:\(x\ge0\)

\(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)=7\\ \Leftrightarrow2x+\sqrt{x}-4\sqrt{x}-2=7\\ \Leftrightarrow2x-3\sqrt{x}-9=0\\ \Leftrightarrow\left(2x+3\sqrt{x}\right)-\left(6\sqrt{x}+9\right)=0\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+3\right)-3\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left(\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\2\sqrt{x}=-3\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=9\left(tm\right)\)

\(< =>-2\sqrt{x}=-4\\ < =>\sqrt{x}=2\\ < =>x=4\)

ĐKXĐ: x>=0

\(3-2\sqrt{x}=-1\)

=>\(2\sqrt{x}=3+1=4\)

=>\(\sqrt{x}=2\)

=>x=4(nhận)

Nếu chứng minh $\sqrt{x}+\sqrt{x+1}=1$ thì không có đủ cơ sở để cm bạn nhé. Bạn viết lại đề hoặc bổ sung thêm điều kiện để mọi người trợ giúp tốt hơn.