a)\(4x+3x=217\)
b)\(9x-3x=216\)
c)\(6x-3x+23=230\)
d)\(5x+3x+x=72\)
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Những câu hỏi liên quan
25 tháng 5 2022
a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)
=>-38x=7
hay x=-7/38
b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)
=>1/2x=0
hay x=0
c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)
=>-29x=15
hay x=-15/29
d: \(\Leftrightarrow x^2+2x-x-3=5\)
\(\Leftrightarrow x^2+x-8=0\)
\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)
e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)
\(\Leftrightarrow-25x^2=4\)
\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)
25 tháng 3 2020
a) Ta có: \(3x^2\cdot\left(2x^3-x+5\right)\)
\(=6x^5-3x^3+15x^2\)
b) Ta có: \(\left(4xy+3y-5\right)\cdot x^2y\)
\(=4x^3y^2+3x^2y^2-5x^2y\)
c) Ta có: \(\left(3x-2\right)\left(4x+5\right)-6x\left(2x-1\right)\)
\(=12x^2+15x-8x-10-12x^2+6x\)
\(=13x-10\)
25 tháng 3 2020
d) Ta có: \(\left(3x-5\right)\left(x^2-5x+7\right)\)
\(=3x^3-15x^2+21x-5x^2+25x-35\)
\(=3x^3-20x^2+46x-35\)
31 tháng 5 2021
\(A=-x^2+3x-5\)\(=-\dfrac{11}{4}-\left(x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}\right)=-\dfrac{11}{4}-\left(x-\dfrac{3}{2}\right)^2\le-\dfrac{11}{4}\) với mọi x
\(\Rightarrow A_{max}=-\dfrac{11}{4}\Leftrightarrow x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)
\(B=5x-4x^2-3=-\dfrac{23}{16}-\left(4x^2-2.\dfrac{5}{4}.2x+\dfrac{25}{16}\right)\)\(=-\dfrac{23}{16}-\left(2x-\dfrac{5}{4}\right)^2\)\(\le-\dfrac{23}{16}\forall x\)
\(\Rightarrow B_{max}=-\dfrac{23}{16}\Leftrightarrow2x-\dfrac{5}{4}=0\Leftrightarrow x=\dfrac{5}{8}\)
\(C=5-4x-25x^2=\dfrac{129}{25}-\left(25x^2+2.5x.\dfrac{2}{5}+\dfrac{4}{25}\right)\)\(=\dfrac{129}{25}-\left(5x+\dfrac{2}{5}\right)^2\le\dfrac{129}{25}\forall x\)
\(\Rightarrow C_{max}=\dfrac{129}{25}\Leftrightarrow5x+\dfrac{2}{5}=0\Leftrightarrow x=-\dfrac{2}{25}\)
31 tháng 5 2021
\(D=3x-2x^2=-2\left(x^2-\dfrac{3}{2}x\right)=-2\left(x^2-2.\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{9}{8}\)\(=\dfrac{9}{8}-2\left(x-\dfrac{3}{4}\right)^2\le\dfrac{9}{8}\) với mọi x
\(\Rightarrow D_{max}=\dfrac{9}{8}\Leftrightarrow x-\dfrac{3}{4}=0\Leftrightarrow x=\dfrac{3}{4}\)
\(E=2+6x-\dfrac{1}{4}x^2=-\dfrac{1}{4}\left(x^2-24x\right)+2=-\dfrac{1}{4}\left(x^2-2.12x+144\right)+38\)\(=38-\dfrac{1}{4}\left(x-12\right)^2\le38\forall x\)
\(\Rightarrow E_{max}=38\Leftrightarrow x-12=0\Leftrightarrow x=12\)
\(F=-5x^2+4x=-5\left(x^2-\dfrac{4}{5}x\right)=-5\left(x^2-2.\dfrac{2}{5}x+\dfrac{4}{25}\right)+\dfrac{4}{5}\)\(=\dfrac{4}{5}-5\left(x-\dfrac{2}{5}\right)^2\le\dfrac{4}{5}\forall x\)
\(\Rightarrow F_{max}=\dfrac{4}{5}\Leftrightarrow x-\dfrac{2}{5}=0\Leftrightarrow x=\dfrac{2}{5}\)
YK
3
H
5 tháng 2 2023
\(b,x^2+3x-2=0\\ \Delta=3^2-4.1.\left(-2\right)=17\\ =>\left[{}\begin{matrix}x_1=\dfrac{-3+\sqrt{17}}{2}\\x_2=\dfrac{-3-\sqrt{17}}{2}\end{matrix}\right.\)
Mấy câu còn lại mình giải rồi
5 tháng 7 2018
1) a) \(\left(3x-1\right)\left(9x^2+3x+1\right)-4x\left(x-5\right)\)
\(=27x^3+9x^2+3x-9x^2-3x-1-4x^2+20x\)
\(=27x^3+\left(9x^2-9x^2-4x^2\right)+\left(3x-3x+20x\right)+\left(-1\right)\)
\(=27x^3-4x^2+20x-1\)
b)\(\left(7x+2\right)\left(3-4x\right)-\left(x+3\right)\left(x^2-3x+9\right)\)
\(=21x-28x^2+6-8x-x^3+3x^2-9x-3x^2+9x-27\)
\(=\left(21x-8x-9x+9x\right)+\left(-28x^2+3x^2-3x^2\right)\)\(+\left(6-27\right)\)\(+\left(-x^3\right)\)
\(=13x-28x^2-21-x^3\)
c)\(\left(4x+3\right)\left(4x-3\right)-\left(2-x\right)\left(4+2x+x^2\right)\)
\(=16x^2-12x+12x-9-8-4x-2x^2+4x+2x^2+x^3\)
\(=\left(16x^2-2x^2+2x^2\right)+\left(-12x+12x-4x+4x\right)\)\(+\left(-9-8\right)\)\(+x^3\)
\(=16x^2-17+x^3\)
d)\(\left(3x-8\right)\left(-5x+6\right)-\left(4x+1\right)\left(3x-2\right)\)
\(=-15x^2+18x+40x-48-12x^2+8x-3x+2\)
\(=\left(-15x^2-12x^2\right)+\left(18x+40x+8x-3x\right)\)\(+\left(-48+2\right)\)
\(=-27x^2+63x-46\)
e)\(\left(3x-6\right)4x-2x\left(3x+5\right)-4x^2\)
\(=12x^2-24x-6x^2-10x-4x^2\)
\(=\left(12x^2-6x^2-4x^2\right)+\left(-24x-10x\right)\)
\(=2x^2-34x\)
f)\(\left(5x-6\right)\left(6x-5\right)-x\left(3x+10\right)\)
\(=30x^2-25x-36x+30-3x^2-10x\)
\(=\left(30x^2-3x^2\right)+\left(-25x-36x-10x\right)+30\)
\(=27x^2-71x+30\)
5 tháng 7 2018
2) a)\(x\left(x+3\right)-x^2=6\)
\(\Rightarrow x^2+3x-x^2=6\)
\(\Rightarrow\left(x^2-x^2\right)+3x=6\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
Vậy x=2
b) \(2x\left(x-5\right)+x\left(-2x-1\right)=6\)
\(\Rightarrow2x^2-10x-2x^2-x=6\)
\(\Rightarrow\left(2x^2-2x^2\right)+\left(-10x-x\right)=6\)
\(\Rightarrow-11x=6\)
\(\Rightarrow x=-\dfrac{6}{11}\)
\(\)Vậy \(x=-\dfrac{6}{11}\)
c) x(x+5)-(x+1)(x-2)=7
\(\Rightarrow x^2+5x-x^2+2x-x+2=7\)
\(\Rightarrow\left(x^2-x^2\right)+\left(5x+2x-x\right)=7-2\)
\(\Rightarrow6x=5\)
\(\Rightarrow x=\dfrac{5}{6}\)
Vậy x=\(\dfrac{5}{6}\)
d)\(\left(3x+4\right)\left(6x-3\right)-\left(2x+1\right)\left(9x-2\right)=10\)
\(\Rightarrow18x^2-9x+24x-12-18x^2+4x-9x+2=10\)
\(\Rightarrow\left(18x^2-18x^2\right)+\left(-9x+24x+4x-9x\right)+\left(-12+2\right)=10\)
\(\Rightarrow10x-10=10\)
\(\Rightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy x=2
MH
18 tháng 10 2023
a) \(x^2-5x+6\)
\(=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
b) \(x^2-9x+18=x^2-3x-6x+18\)
\(=x\left(x-3\right)-6\left(x-3\right)=\left(x-3\right)\left(x-6\right)\)
c) \(x^2-6x+5=x^2-x-5x+5\)
\(=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)
d) \(3x^2+5x-30=3\left(x^2+\dfrac{5x}{3}-10\right)=3\left(x^2+2.x.\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{5347}{500}\right)\)
Câu này bạn xem lại đề nha
e) \(3x^2-5x-2=3x^2-6x+x-2\)
\(3x\left(x-2\right)+x-2=\left(x-2\right)\left(3x+1\right)\)
C
3
25 tháng 6 2021
`A=x^2-2x+5`
`=x^2-2x+1+4`
`=(x-1)^2+4>=4`
Dấu "=" `<=>x=1`
`B=4x^2+4x+3`
`=4x^2+4x+1+2`
`=(2x+1)^2+2>=2`
Dấu "=" xảy ra khi `x=-1/2`
`C=9x^2-6x+7`
`=9x^2-6x+1+6`
`=(3x-1)^2+6>=6`
Dấu '=' xảy ra khi `x=1/3`
`D=5x^2+3x+8`
`=5(x^2+3/5x)+8`
`=5(x^2+3/5x+9/100-9/100)+8`
`=5(x+3/10)^2+151/20>=151/20`
Dấu "=" xảy ra khi `x=-3/10`
AT
25 tháng 6 2021
\(A=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\)
Ta có: \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+4\ge4\Rightarrow A_{min}=4\) khi \(x=1\)
\(B=4x^2+4x+3=4x^2+4x+1+2=\left(2x+1\right)^2+2\)
Ta có: \(\left(2x+1\right)^2\ge0\Rightarrow\left(2x+1\right)^2+2\ge2\Rightarrow B_{min}=2\) khi \(x=-\dfrac{1}{2}\)
\(C=9x^2-6x+7=9x^2-6x+1+6=\left(3x-1\right)^2+6\)
Ta có: \(\left(3x-1\right)^2\ge0\Rightarrow\left(3x-1\right)^2+6\ge6\Rightarrow C_{min}=6\) khi \(x=\dfrac{1}{3}\)
\(D=5x^2+3x+8\Rightarrow5\left(x^2+2.x.\dfrac{3}{10}+\dfrac{9}{100}\right)+\dfrac{151}{20}=5\left(x+\dfrac{3}{10}\right)^2+\dfrac{151}{20}\)
Ta có: \(5\left(x+\dfrac{3}{10}\right)^2\ge0\Rightarrow5\left(x+\dfrac{3}{10}\right)^2+\dfrac{151}{20}\ge\dfrac{151}{20}\)
\(\Rightarrow D_{min}=\dfrac{151}{20}\) khi \(x=-\dfrac{3}{10}\)
a) 4x + 3x = 217
x( 4 + 3 ) = 217
7x = 217
x = 217 : 7 = 31
Vậy x = 31
b) 9x - 3x = 216
( 9 -3)x = 216
6x = 216
x = 216:6 = 36
Vậy x = 36
c) 6x - 3x + 23 = 230
( 6 - 3 )x = 230 - 23
3x = 207
x = 207 : 3 = 69
Vậy x = 69
d) 5x + 3x + x = 72
5x + 3x + 1x = 72
( 5 + 3 + 1 )x = 72
9x = 72
x = 72 : 9 = 8
Vậy x = 8
Chúc bạn học tốt nhé
a) \(4x+3x=217\)
\(\Rightarrow x\cdot\left(3+4\right)=217\)
\(\Rightarrow7x=217\)
\(\Rightarrow x=\dfrac{217}{7}\)
\(\Rightarrow x=31\)
b) \(9x-3x=216\)
\(\Rightarrow x\cdot\left(9-3\right)=216\)
\(\Rightarrow6x=216\)
\(\Rightarrow x=\dfrac{216}{6}\)
\(\Rightarrow x=36\)
c) \(6x-3x+23=230\)
\(\Rightarrow x\cdot\left(6-3\right)=230-23\)
\(\Rightarrow3x=207\)
\(\Rightarrow x=\dfrac{207}{3}\)
\(\Rightarrow x=69\)
d) \(5x+3x+x=72\)
\(\Rightarrow x\cdot\left(5+3+1\right)=72\)
\(\Rightarrow9x=72\)
\(\Rightarrow x=\dfrac{72}{9}\)
\(\Rightarrow x=8\)