CMR
f(x)= 10x+5
g(x)=\(x^3+\dfrac{1}{2}x^2+3x+\dfrac{3}{2}\). Có nghiệm chung
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TT
1
AH
Akai Haruma
Giáo viên
4 tháng 8 2018
Lời giải:
\(f(x)=2x+1\)
\(g(x)=x^3+\frac{1}{2}x^2+3x+\frac{3}{2}\)
\(=x^2(x+\frac{1}{2})+3(x+\frac{1}{2})\)
\(=(x^2+3)(x+\frac{1}{2})=\frac{1}{2}(x^2+3)(2x+1)\)
Do đó $f(x),g(x)$ có chung nhân tử \(2x+1\) nên có chung nghiệm \(x=-\frac{1}{2}\)
11 tháng 3 2023
f: =>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
=>28-6(x+2)=-9-5(x-4)
=>28-6x-12=-9-5x+20
=>-6x+16=-5x+11
=>-x=-5
=>x=5
12 tháng 3 2023
d: =>\(\dfrac{12x+1}{11x-4}=\dfrac{20x+17-20x+8}{18}=\dfrac{25}{18}\)
=>25(11x-4)=18(12x+1)
=>275x-100=216x+18
=>59x=118
=>x=2
f: =>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
=>28-6(x+2)=-9-5(x-4)
=>28-6x-12=-9-5x+20
=>-6x+16=-5x+11
=>-x=-5
=>x=5
5 tháng 2 2022
e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)
\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)
\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)
\(\Leftrightarrow x=-1\left(TM\right)\)
24 tháng 2 2021
a) Ta có: \(2\left(3x+1\right)-4\left(5-2x\right)>2\left(4x-3\right)-6\)
\(\Leftrightarrow6x+2-20+8x>8x-6-6\)
\(\Leftrightarrow14x-18-8x+12>0\)
\(\Leftrightarrow6x-6>0\)
\(\Leftrightarrow6x>6\)
hay x>1
Vậy: S={x|x>1}
b) Ta có: \(9x^2-3\left(10x-1\right)< \left(3x-5\right)^2-21\)
\(\Leftrightarrow9x^2-30x+3< 9x^2-30x+25-21\)
\(\Leftrightarrow9x^2-30x+3-9x^2+30x-4< 0\)
\(\Leftrightarrow-1< 0\)(luôn đúng)
Vậy: S={x|\(x\in R\)}
21 tháng 1 2021
2) Ta có: \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
\(\Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30\left(x-4\right)}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)
\(\Leftrightarrow6x+24-30x+120=10x-15x+30\)
\(\Leftrightarrow-24x+144=-5x+30\)
\(\Leftrightarrow-24x+144+5x-30=0\)
\(\Leftrightarrow-19x+114=0\)
\(\Leftrightarrow-19x=-114\)
hay x=6
Vậy: x=6
3) Ta có: \(\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
\(\Leftrightarrow\dfrac{3\left(10x+3\right)}{36}=\dfrac{36}{36}+\dfrac{4\left(6+8x\right)}{36}\)
\(\Leftrightarrow30x+9=36+24+32x\)
\(\Leftrightarrow30x+9-60-32x=0\)
\(\Leftrightarrow-2x-51=0\)
\(\Leftrightarrow-2x=51\)
hay \(x=-\dfrac{51}{2}\)
Vậy: \(x=-\dfrac{51}{2}\)
4) Ta có: \(\dfrac{x+1}{3}-\dfrac{x-2}{6}=\dfrac{2x-1}{2}\)
\(\Leftrightarrow\dfrac{2\left(x+1\right)}{6}-\dfrac{x-2}{6}=\dfrac{3\left(2x-1\right)}{6}\)
\(\Leftrightarrow2x+2-x+2=6x-3\)
\(\Leftrightarrow x+4-6x+3=0\)
\(\Leftrightarrow-5x+7=0\)
\(\Leftrightarrow-5x=-7\)
hay \(x=\dfrac{7}{5}\)
Vậy: \(x=\dfrac{7}{5}\)
KV
21 tháng 1 2021
1) \(\dfrac{5x-2}{3}=\dfrac{5-3x}{2}\)
\(2\left(5x-2\right)=3\left(5-3x\right)\)
\(10x-4=15-9x\)
\(10x+9x=15+4\)
\(19x=19\)
\(x=1\)
Vậy \(x=1\)
8 tháng 3 2021
a) ĐKXĐ: \(x\notin\left\{-1;-2;2\right\}\)
Ta có: \(\dfrac{1}{x^2+3x+2}-\dfrac{3}{x^2-x-2}=\dfrac{-1}{x^2-4}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}-\dfrac{3}{\left(x-2\right)\left(x+1\right)}=\dfrac{-1}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{x-2}{\left(x+1\right)\left(x+2\right)\left(x-2\right)}-\dfrac{3\left(x+2\right)}{\left(x+2\right)\left(x+1\right)\left(x-2\right)}=\dfrac{-1\left(x+1\right)}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x-2-3x-6=-x-1\)
\(\Leftrightarrow-2x-8+x+1=0\)
\(\Leftrightarrow-x-7=0\)
\(\Leftrightarrow-x=7\)
hay x=-7(thỏa ĐK)
Vậy: S={-7}
TP
8 tháng 3 2021
a) ĐKXĐ: x∉{−1;−2;2}x∉{−1;−2;2}
Ta có: 1x2+3x+2−3x2−x−2=−1x2−41x2+3x+2−3x2−x−2=−1x2−4
⇔1(x+1)(x+2)−3(x−2)(x+1)=−1(x−2)(x+2)⇔1(x+1)(x+2)−3(x−2)(x+1)=−1(x−2)(x+2)
⇔x−2(x+1)(x+2)(x−2)−3(x+2)(x+2)(x+1)(x−2)=−1(x+1)(x+1)(x−2)(x+2)⇔x−2(x+1)(x+2)(x−2)−3(x+2)(x+2)(x+1)(x−2)=−1(x+1)(x+1)(x−2)(x+2)
Suy ra: x−2−3x−6=−x−1x−2−3x−6=−x−1
⇔−2x−8+x+1=0⇔−2x−8+x+1=0
⇔−x−7=0⇔−x−7=0
⇔−x=7⇔−x=7
hay x=-7(thỏa ĐK)
Vậy: S={-7}
Đọc tiếp
a) ĐKXĐ: x∉{−1;−2;2}x∉{−1;−2;2}
Ta có: 1x2+3x+2−3x2−x−2=−1x2−41x2+3x+2−3x2−x−2=−1x2−4
⇔1(x+1)(x+2)−3(x−2)(x+1)=−1(x−2)(x+2)⇔1(x+1)(x+2)−3(x−2)(x+1)=−1(x−2)(x+2)
⇔x−2(x+1)(x+2)(x−2)−3(x+2)(x+2)(x+1)(x−2)=−1(x+1)(x+1)(x−2)(x+2)⇔x−2(x+1)(x+2)(x−2)−3(x+2)(x+2)(x+1)(x−2)=−1(x+1)(x+1)(x−2)(x+2)
Suy ra: x−2−3x−6=−x−1x−2−3x−6=−x−1
⇔−2x−8+x+1=0⇔−2x−8+x+1=0
⇔−x−7=0⇔−x−7=0
⇔−x=7⇔−x=7
hay x=-7(thỏa ĐK)
Vậy: S={-7}
Đọc tiếp
NT
1
8 tháng 9 2022
\(\dfrac{x}{x^2+x+1}=\dfrac{1}{4}\)
=>\(x^2+x+1=4x\)
=>\(x^2-3x+1=0\)
\(F=\dfrac{x^5-3x^4+x^3+3x^4-9x^3+3x^2+5x^3-15x^2+5x+12x^2-36x+12+21x}{x^2\left(x^2-3x+1\right)+3x\left(x^2-3x+1\right)+15\left(x^2-3x+1\right)+27x}\)
\(=\dfrac{12x}{27x}=\dfrac{4}{9}\)
\(f\left(x\right)=10x+5\)
Nghiệm của \(f\left(x\right)\) là \(x\Leftrightarrow10x+5=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\) (1)
\(g\left(x\right)=x^3+\dfrac{1}{2}x^2+3x+\dfrac{3}{2}\)
Nghiệm của \(g\left(x\right)\)là \(x\Leftrightarrow x^3+\dfrac{1}{2}x^2+3x+\dfrac{3}{2}=0\)
\(\Leftrightarrow x^2\left(x+\dfrac{1}{2}\right)+3\left(x+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)\left(x^2+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\x^2+3=0\end{matrix}\right.\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\) (2)
Từ (1) và (2) => ĐPCM