\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)⇒\(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b-3c}{2+6-12}=\dfrac{-20}{-4}=5\)

⇒\(\left\{{}\begin{matrix}x=5.2=10\\y=5.3=15\\z=5.4=20\end{matrix}\right.\)

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}\)

Áp dụng t/c của DS bằng nhau, ta có: \(\dfrac{a+2b-3c}{2+6-12}=\dfrac{-20}{-4}=5\)

\(\dfrac{a}{2}=5\Rightarrow a=10\)

\(\dfrac{b}{3}=5\Rightarrow b=15\)

\(\dfrac{c}{4}=5\Rightarrow c=20\)

tham khảo:

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{a+2b-3c}{2+2x3-3x4}=5\)

\(\Rightarrow\dfrac{a}{2}=5\Rightarrow a=5x2=10\)

\(\Rightarrow\dfrac{b}{3}=5\Rightarrow b=5x3=15\)

\(\Rightarrow\dfrac{c}{4}=5\Rightarrow c=20\)

\(\left(12x+7\right)^2\left(3x+2\right)\left(2x+1\right)=3\)

\(\Leftrightarrow\left(144x^2+168x+49\right)\left(6x^2+7x+2\right)=3\)

\(\Leftrightarrow\left(144x^2+168x+49\right).24.\left(6x^2+7x+2\right)=3.24\)

\(\Leftrightarrow\left(144x^2+168x+49\right)\left(144x^2+168x+48\right)=72\) (*)

-Đặt \(144x^2+168x+49=a\)

(*) \(\Leftrightarrow a\left(a-1\right)=72\)

\(\Leftrightarrow a^2-a-72=0\)

\(\Leftrightarrow a^2-9a+8a-72=0\)

\(\Leftrightarrow a\left(a-9\right)+8\left(a-9\right)=0\)

\(\Leftrightarrow\left(a-9\right)\left(a+8\right)=0\)

\(\Leftrightarrow a-9=0\) hay \(a+8=0\)

\(\Leftrightarrow a=9\) hay \(a=-8\)

\(\Leftrightarrow144x^2+168x+40=0\) hay \(144x^2+168x+57=0\)

\(\Leftrightarrow144x^2+48x+120x+40=0\) hay ​\(144x^2+168x+49+8=0\)​

\(\Leftrightarrow48x\left(\dfrac{1}{3}x+1\right)+40\left(\dfrac{1}{3}x+1\right)=0\) hay \(\left(12x+7\right)^2+8=0\) (vô nghiệm vì \(\left(12x+7\right)^2+8\ge8\))

\(\Leftrightarrow8\left(\dfrac{1}{3}x+1\right)\left(6x+5\right)=0\)

\(\Leftrightarrow\dfrac{1}{3}x+1=0\) hay \(6x+5=0\)

\(\Leftrightarrow x=\dfrac{-1}{3}\) hay \(x=\dfrac{-5}{6}\)

-Vậy \(S=\left\{\dfrac{-1}{3};\dfrac{-5}{6}\right\}\)

\(144x^2+168x+49=a\)

90 phút nữa mk lm :)

bài3:giá trị biểu thức

a)824x4+1043

 =3296+1043

=4339

b)6743-2640+345

=4103+345

=4448

a) 824x4+1043= 3296+1043= 4339

b) 6743-2640+345= 4103+4448

a)4339

b)4448

Ta có phương trình phản ứng hóa học sau 2K + 2H2O -----> 2KOH + H2

Số mol của K = 3,9/ 39 = 0,1

=> mol KOH = 0,1

=> mol H2O = 0,05

Theo cân bằng phương trình ta tính được:

mdd = mk + mH2O - mH2 = 3,9 + 36,2 - (0,05.2) = 40 gam

=> Áp dụng công thức C% = (mct/mdd).100%

ta có C% = [(0,1.56)/40].100% = 14%

\(\left(m-2\right)x^4-2\left(m+1\right)x^2+2m-1=0\left(1\right)\)

\(m=2\left(ktm\right)\)

\(m\ne2:đặt:x^2=t\ge0\Rightarrow\left(1\right)\Leftrightarrow\left(m-2\right)t^2-2\left(m+1\right)t+2m-1=0\)

\(3nghiem\Leftrightarrow\left\{{}\begin{matrix}2m-1=0\\t1+t2=\dfrac{2m+2}{m-2}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=\dfrac{1}{2}\\\left[{}\begin{matrix}m>2\\m< -1\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow m\in\phi\)

\(4nghiem\Leftrightarrow\left\{{}\begin{matrix}\Delta'>0\\t1+t2>0\Leftrightarrow\\t1.t2>0\end{matrix}\right.\left\{{}\begin{matrix}\left(m+1\right)^2-\left(m-2\right)\left(2m-1\right)>0\\\dfrac{2m+2}{m-2}>0\\\dfrac{2m-1}{m-2}>0\end{matrix}\right.\)

giải hệ bất pt trên=>m

\(c3:b;\left\{{}\begin{matrix}-8\le x\le-2\\m\left(x-3\right)\ge1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-8\le x\le-2\\m\le\dfrac{1}{x-3}\end{matrix}\right.\)

\(có\) \(nghiệm\Leftrightarrow m\le max:\dfrac{1}{x-3}trên\left[-8;-2\right]\)

\(\Leftrightarrow m\le\dfrac{-1}{5}\)

Theo công thức S = \(\frac{\pi R^2n^o}{360^0}\)  ta có S=\(\frac{\pi6^2.36}{360}\)  \(\approx\) 3,6\(\pi\) ( cm\(^2\))

                                            @hc tốt !!

                                          #Ji_en *quẹc*