Bài 2. (2điểm)Tìm x biếta) 3(2x –3) + 2(2 –x) = –3;b) (x –2021)(x –5) = x –2021;c) (2x –3)2–36x2=...
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Bài 2. (2điểm)Tìm x biếta) 3(2x –3) + 2(2 –x) = –3;b) (x –2021)(x –5) = x –2021;c) (2x –3)2–36x2= 0.
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NL
1
27 tháng 9 2021
a: Ta có: \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)
\(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)
\(\Leftrightarrow-12x=24\)
hay x=-2
b: Ta có: \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)
\(\Leftrightarrow2x=-40\)
hay x=-20
27 tháng 6 2023
2:
a: A(x)=0
=>5x-10-2x-6=0
=>3x-16=0
=>x=16/3
b: B(x)=0
=>5x^2-125=0
=>x^2-25=0
=>x=5 hoặc x=-5
c: C(x)=0
=>2x^2-x-3=0
=>2x^2-3x+2x-3=0
=>(2x-3)(x+1)=0
=>x=3/2 hoặc x=-1
19 tháng 8 2023
a: =>1/3(2x-5)=-2/3-3/2=-4/6-9/6=-13/6
=>2x-5=-13/6*3=-13/2
=>2x=-3/2
=>x=-3/4
b: =>2/5x=-3/4-1/2=-5/4
=>x=-5/4:2/5=-5/4*5/2=-25/8
GH
19 tháng 8 2023
a)
\(-\dfrac{2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}\\ \Rightarrow\dfrac{1}{3}\left(2x-5\right)=-\dfrac{2}{3}-\dfrac{3}{2}\\ \Rightarrow\dfrac{1}{3}\left(2x-5\right)=-\dfrac{13}{6}\\ \Rightarrow2x-5=-\dfrac{13}{6}:\dfrac{1}{3}=-\dfrac{13}{2}\\ \Rightarrow2x=-\dfrac{13}{2}+5\\ \Rightarrow2x=-\dfrac{3}{2}\\ \Rightarrow x=-\dfrac{3}{2}:2\\ \Rightarrow x=-\dfrac{3}{4}\)
b)
\(\dfrac{2}{5}x+\dfrac{1}{2}=-\dfrac{3}{4}\\ \Rightarrow\dfrac{2}{5}x=-\dfrac{3}{4}-\dfrac{1}{2}\\ \Rightarrow\dfrac{2}{5}x=-\dfrac{5}{4}\\ \Rightarrow x=-\dfrac{5}{4}:\dfrac{2}{5}=-\dfrac{25}{8}\)
26 tháng 9 2021
e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
f: Ta có: \(x^3-6x^2+12x-19=0\)
\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)
\(\Leftrightarrow\left(x-2\right)^3=11\)
hay \(x=\sqrt[3]{11}+2\)
NN
1
H
26 tháng 12 2022
\(B1\\ a,2x+10y=2\left(x+5y\right)\\ b,x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\\ c,x^2-y^2+10y-25\\ =\left(x^2-y^2\right)+5\left(2y-5\right)\\ =\left(x-y\right)\left(x+y\right)+5\left(2y-5\right)\\ B2\)
\(a,x^2-3x+x-3=0\\ =>x\left(x-3\right)+\left(x-3\right)=0\\ =>\left(x+1\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ b,2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\\ =>2x^2-6x-2x^2+\dfrac{3}{2}=0\\ =>-6x=-\dfrac{3}{2}\\ =>x=\left(-\dfrac{3}{2}\right):\left(-6\right)\\ =>x=\dfrac{1}{4}\\ c,x^2-\left(x-3\right)\left(2x-5\right)=9\\ =>x^2-2x^2+6x+5x-15=9\\ =>-x^2+11-15-9=0\\ =>-x^2+11x-24=0\\ =>-x^2+8x+3x-24=0\\ =>-x\left(x-8\right)+3\left(x-8\right)=0\\ =>\left(3-x\right)\left(x-8\right)=0\\ =>\left[{}\begin{matrix}3-x=0\\x-8=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)
12 tháng 7 2021
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
QN
2
TA
22 tháng 7 2021
a) `4\sqrt(2x-1)>8`
`<=>\sqrt(2x-1)>2`
`<=>2x-1>4`
`<=>x>5/2`
b) `2\sqrtx-1>3`
`<=>2\sqrtx>4`
`<=>\sqrtx>2`
`<=>x>4`
22 tháng 7 2021
a) Ta có: \(4\sqrt{2x-1}>8\)
\(\Leftrightarrow2x-1>4\)
\(\Leftrightarrow2x>5\)
hay \(x>\dfrac{5}{2}\)
b) Ta có: \(2\sqrt{x}-1>3\)
\(\Leftrightarrow\sqrt{x}>2\)
hay x>4
\(a,\Leftrightarrow6x-9+4-2x=-3\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\\ b,\Leftrightarrow\left(x-2021\right)\left(x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2021\\x=6\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-3-6x\right)\left(2x-3+6x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-3-4x=0\\8x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{8}\end{matrix}\right.\)