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a) \(7+2x=22-3x\)
\(\Leftrightarrow5x=15\Leftrightarrow x=3\)
b) \(8x-3=5x+12\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
c) \(x-12+4x=25+2x-1\)
\(\Leftrightarrow3x=36\Leftrightarrow x=12\)
d) \(x+2x+3x-19=3x+5\)
\(\Leftrightarrow3x=24\Leftrightarrow x=8\)
e) \(7-\left(2x+4\right)=-\left(x+4\right)\)
\(\Leftrightarrow x=7\)
f) \(\left(x-1\right)-\left(2x-1\right)=9-x\)
\(\Leftrightarrow0x=9\) ( vô lí )
a: \(\Leftrightarrow4\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3x^2\)
\(\Leftrightarrow4\cdot\left[\left(x^2+60\right)^2+33x\left(x^2+60\right)+272x^2\right]=3x^2\)
=>4(x^2+60)^2+132x(x^2+60)+1085x^2=0
=>4(x^2+60)^2+62x(x^2+60)+70x(x^2+60)+1085x^2=0
=>2(x^2+60)(2x^2+120+31x)+35x(2x^2+120+31x)=0
=>(2x^2+120+35x)(2x^2+31x+120)=0
=>\(x\in\left\{\dfrac{-35\pm\sqrt{265}}{4};-\dfrac{15}{2};-8\right\}\)
b: Đặt x^2-3x=a
Phương trình sẽ là \(\dfrac{1}{a+3}+\dfrac{2}{a+4}=\dfrac{6}{a+5}\)
\(\Leftrightarrow\dfrac{a+4+2a+6}{\left(a+3\right)\left(a+4\right)}=\dfrac{6}{a+5}\)
=>(3a+10)(a+5)=6(a^2+7a+12)
=>6a^2+42a+72=3a^2+15a+10a+50
=>3a^2+17a+22=0
=>x=-2 hoặc x=-11/3
\(\frac{4}{2x+3}-\frac{7}{3x-5}=0\left(đkxđ:x\ne-\frac{3}{2};\frac{5}{3}\right)\)
\(< =>\frac{4\left(3x-5\right)}{\left(2x+3\right)\left(3x-5\right)}-\frac{7\left(2x+3\right)}{\left(2x+3\right)\left(3x-5\right)}=0\)
\(< =>12x-20-14x-21=0\)
\(< =>2x+41=0< =>x=-\frac{41}{2}\left(tm\right)\)
\(\frac{4}{2x-3}+\frac{4x}{4x^2-9}=\frac{1}{2x+3}\left(đk:x\ne-\frac{3}{2};\frac{3}{2}\right)\)
\(< =>\frac{4\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}+\frac{4x}{\left(2x-3\right)\left(2x+3\right)}-\frac{2x-3}{\left(2x+3\right)\left(2x-3\right)}=0\)
\(< =>8x+12+4x-2x+3=0\)
\(< =>10x=15< =>x=\frac{15}{10}=\frac{3}{2}\left(ktm\right)\)
\(a,2x-6< 0\Leftrightarrow2x>6\Leftrightarrow x>3\)
\(b,5x+2x< 4+25\Leftrightarrow7x< 29\Leftrightarrow x< \frac{29}{7}\)
\(c,-5x+6>8-10+8x\Leftrightarrow-5x-8x>8-10-6\)
\(-13x>-8\Leftrightarrow x< \frac{8}{13}\)
\(d,3x-12\le2-4x\Leftrightarrow3x+4x\le2+12\)
\(\Leftrightarrow7x\le14\Leftrightarrow x\le2\)
\(e,\frac{3\left(x-3\right)}{6}>\frac{2\left(2x-5\right)}{6}+\frac{6}{6}\Rightarrow3x-9>4x-10+6\)
\(\Leftrightarrow3x-4x>-4+9\Leftrightarrow x>-5\)
\(f,3\left(2x-3\right)>1+2\left(2+2x\right)\Leftrightarrow6x-9>1+4+4x\)
\(6x-4x>14\Leftrightarrow2x>14\Leftrightarrow x>7\)
Tự biểu diễn nha!
a) 4 ( x + 5 )( x + 6 )( x + 10 )( x + 12 ) = 3x2
Do x = 0 không là nghiệm pt nên chia 2 vế pt cho \(x^2\ne0\), ta được :
\(\frac{4}{x^2}\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3\)
\(\Leftrightarrow4\left(x+\frac{60}{x}+17\right)\left(x+\frac{60}{x}+16\right)=3\)
Đến đây ta đặt \(x+\frac{60}{x}+16=t\left(1\right)\)
Ta được :
\(4t\left(t+1\right)=3\Leftrightarrow4t^2+4t-3=0\Leftrightarrow\left(2t+3\right)\left(2t-1\right)=0\)
Từ đó ta lắp vào ( 1 ) tính được x
a: \(A=-\left(x^2-4x-3\right)\)
\(=-\left(x^2-4x+4-7\right)\)
\(=-\left(x-2\right)^2+7< =7\)
Dấu '=' xảy ra khi x=2
b: \(B=-\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\right)=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}< =\dfrac{1}{4}\)
Dấu '=' xảy ra khi x=1/2
c: \(C=-2\left(x^2-x+\dfrac{5}{2}\right)\)
\(=-2\left(x^2-x+\dfrac{1}{4}+\dfrac{9}{4}\right)\)
\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}< =-\dfrac{9}{2}\)
Dấu '=' xảy ra khi x=1/2
e: \(E=-\left(x^2+6x+9+1\right)=-\left(x+3\right)^2-1< =-1\)
Dấu = xảy ra khi x=-3
4 x 2 – 25 – (2x + 7)(5 – 2x)
= ( 2 x ) 2 – 5 2 – (2x + 7)(5 – 2x)
= (2x – 5)(2x + 5) – (2x + 7)(5 – 2x)
= (2x- 5)(2x + 5) + (2x + 7)(2x – 5)
= (2x – 5)(2x + 5 + 2x + 7)
= (2x – 5)(4x + 12)
Biểu thức cần điền là 4x + 12
Đáp án cần chọn là: D