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DH
12 tháng 6 2017
a, \(C=5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
\(C=5.\left(4x^2-4x+1\right)+4\left(x^2+3x-x-3\right)-2.\left(25-75x+9x^2\right)\)
\(C=20x^2-20x+5+4x^2+8x-12-50+150x-18x^2\)
\(=\left(20x^2+4x^2-18x^2\right)+\left(-20x+8x+150x\right)+\left(5-12-50\right)\)
\(C=6x^2+138x-57\)
Chúc bạn học tốt!!! Cũng không chắc có đúng hay sai nữa do cồng kềnh quá !
29 tháng 7 2021
Bài 209 : đăng tách ra cho mn cùng làm nhé
a,sửa đề : \(A=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left(3x+1-3x-5\right)^2=\left(-4\right)^2=16\)
b, \(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)\)
\(2B=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)=\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(2B=3^{64}-1\Rightarrow B=\frac{3^{64}-1}{2}\)
c, \(C=\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)
\(=2\left(a-b+c\right)^2-2\left(b-c\right)^2=2\left[\left(a-b+c\right)^2-\left(b-c\right)^2\right]\)
\(=2\left(a-b+c-b+c\right)\left(a-b+c+b-c\right)=2a\left(a-2b+2c\right)\)
9 tháng 3 2022
\(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)\(\left(đkcđ:x\ne\pm3;x\ne-\frac{1}{2}\right)\)
\(=\left(\frac{\left(x-1\right).\left(x-3\right)+2.\left(x+3\right)-\left(x^2+3\right)}{x^2-9}\right):\left(\frac{2x-1-\left(2x+1\right)}{2x+1}\right)\)
\(=\frac{x^2-4x+3+2x+6-x^2-3}{x^2-9}:\frac{-2}{2x+1}\)
\(=\frac{-2x-6}{x^2-9}.\frac{2x+1}{-2}\)
\(=\frac{-2\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}.\frac{2x+1}{-2}\)
\(=\frac{2x+1}{x-3}\)
b)\(\left|x+1\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x+1=\frac{1}{2}\\x+1=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\left(koTMđkxđ\right)\\x=-\frac{3}{2}\left(TMđkxđ\right)\end{cases}}}\)
thay \(x=-\frac{3}{2}\) vào P tâ đc: \(P=\frac{2x+1}{x-3}=\frac{2.\left(-\frac{3}{2}\right)+1}{-\frac{3}{2}-3}=\frac{4}{9}\)
c)ta có:\(P=\frac{x}{2}\Leftrightarrow\frac{2x+1}{x-3}=\frac{x}{2}\)
\(\Rightarrow2.\left(2x+1\right)=x.\left(x-3\right)\)
\(\Leftrightarrow4x+2=x^2-3x\)
\(\Leftrightarrow x^2-7x-2=0\)
\(\Leftrightarrow x^2-2.\frac{7}{2}+\frac{49}{4}-\frac{57}{4}=0\)
\(\Leftrightarrow\left(x-\frac{7}{2}\right)^2-\frac{57}{4}=0\)
\(\Leftrightarrow\left(x-\frac{7}{2}-\frac{\sqrt{57}}{2}\right).\left(x-\frac{7}{2}+\frac{\sqrt{57}}{2}\right)\)
bạn tự giải nốt nhé!!
d)\(x\in Z;P\in Z\Leftrightarrow\frac{2x+1}{x-3}\in Z\Leftrightarrow\frac{2x-6+7}{x-3}=2+\frac{7}{x-3}\in Z\)
\(2\in Z\Rightarrow\frac{7}{x-3}\in Z\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
bạn tự làm nốt nhé
9 tháng 3 2022
a, \(\left(\dfrac{x^2-4x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}\right):\left(\dfrac{2x-1-2x-1}{2x+1}\right)\)
\(=\dfrac{-2x+6}{\left(x+3\right)\left(x-3\right)}:\dfrac{-2}{2x+1}=\dfrac{-2\left(x-3\right)\left(2x+1\right)}{-2\left(x+3\right)\left(x-3\right)}=\dfrac{2x+1}{x+3}\)
b, \(\left|x+1\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}-1\\x=-\dfrac{1}{2}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(ktmđk\right)\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Thay x = -3/2 ta được \(\dfrac{2\left(-\dfrac{3}{2}\right)+1}{-\dfrac{3}{2}+3}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\)
6 tháng 3 2020
\(\left(x^2-4\right)+\left(8-5.x\right).\left(x+2\right)+4.\left(x-2\right).\left(x+1\right)=0\)
\(\Leftrightarrow x^2-4+8.x+16-5.x^2-10.x+\left(4.x-8\right).\left(x+1\right)=0\)
\(\Leftrightarrow x^2-4+8.x+16-5.x^2-10.x+4.x^2+4.x-8.x-8=0\)
\(\Leftrightarrow0+4-6.x=0\)
\(\Leftrightarrow4-6.x=0\)
\(\Leftrightarrow-6.x=-4\)
\(\Rightarrow x=\frac{2}{3}\)
Vậy x = \(\frac{2}{3}\)
DT
9 tháng 2 2021
(2x2+x-6)+3(2x2+x-3)-9=0
\(\Leftrightarrow\) 2x2 + x - 6 + 6x2 + 3x - 9 - 9 = 0
\(\Leftrightarrow\)2x2 + 6x2 + 3x + x = 6 + 9 + 9
\(\Leftrightarrow\)8x2 + 4x = 24
\(\Leftrightarrow\)8x2 + 4x - 24 = 0
\(\Leftrightarrow\)(x+2)(8x-12) = 0
\(\Leftrightarrow\)x + 2 = 0 hoặc 8x - 12 = 0
1) x + 2 = 0 \(\Leftrightarrow\)x = -2
2)8x - 12 = 0 \(\Leftrightarrow\)8x = 12 \(\Leftrightarrow\)x = \(\frac{12}{8}\)
Vậy Tập nghiệm của phương trình đã cho là S ={ -2 ; \(\frac{12}{8}\)}
Trả lời:
a, \(\frac{4x-5}{x-1}=2+\frac{x}{x-1}\left(đkxđ:x\ne1\right)\)
\(\Leftrightarrow\frac{4x-5}{x-1}=\frac{2\left(x-1\right)+x}{x-1}\)
\(\Rightarrow4x-5=2x-2+x\)
\(\Leftrightarrow4x-5=3x-2\)
\(\Leftrightarrow4x-3x=-2+5\)
\(\Leftrightarrow x=3\left(tm\right)\)
Vậy \(S=\left\{3\right\}\)
b, \(\frac{7}{x+2}=\frac{3}{x-5}\left(đkxđ:x\ne-2;x\ne5\right)\)
\(\Leftrightarrow\frac{7\left(x-5\right)}{\left(x+2\right)\left(x-5\right)}=\frac{3\left(x+2\right)}{\left(x+2\right)\left(x-5\right)}\)
\(\Rightarrow7x-35=3x+6\)
\(\Leftrightarrow7x-3x=6+35\)
\(\Leftrightarrow4x=41\)
\(\Leftrightarrow x=\frac{41}{4}\left(tm\right)\)
Vậy \(S=\left\{\frac{41}{4}\right\}\)
c, \(\frac{2x+5}{2x}-\frac{x}{x+5}=0\left(đkxđ:x\ne0;x\ne-5\right)\)
\(\Leftrightarrow\frac{2x+5}{2x}=\frac{x}{x+5}\)
\(\Leftrightarrow\left(2x+5\right)\left(x+5\right)=2x^2\)
\(\Leftrightarrow2x^2+15x+25=2x^2\)
\(\Leftrightarrow2x^2+15x-2x^2=-25\)
\(\Leftrightarrow15x=-25\)
\(\Leftrightarrow x=\frac{-5}{3}\left(tm\right)\)
Vậy \(S=\left\{\frac{-5}{3}\right\}\)
d, \(\frac{12x+1}{11x-4}+\frac{10x-4}{9}=\frac{20x+17}{18}\left(đkxđ:x\ne\frac{4}{11}\right)\)
\(\Leftrightarrow\frac{18\left(12x+1\right)+2\left(10x-4\right)\left(11x-4\right)}{18\left(11x-4\right)}=\frac{\left(20x+17\right)\left(11x-4\right)}{18\left(11x-4\right)}\)
\(\Rightarrow216x+18+\left(20x-8\right)\left(11x-4\right)=220x^2+107x-68\)
\(\Leftrightarrow216x+18+220x^2-168x+32=220x^2+107x-68\)
\(\Leftrightarrow220x^2+48x+50=220x^2+107x-68\)
\(\Leftrightarrow220x^2+48x-220x^2-107x=-68-50\)
\(\Leftrightarrow-59x=-118\)
\(\Leftrightarrow x=2\left(tm\right)\)
Vậy \(S=\left\{2\right\}\)