- Tìm x biết :
(3x+1)2:(-1/4)=-49/4
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NT
2
9 tháng 10 2017
lớp 8 đến bây giờ đã học hằng đẳng thức chưa ạ ? Để mình đưa bạn hướng giải câu đầu !
câu sau : x^2 -9^2 = 0<=> (x-9)^2=0 <=>x-9=0 <=> x=9
PT
0
22 tháng 9 2021
a: 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 \(x=-\dfrac{7}{2}\)
b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)
\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)
\(\Leftrightarrow24x=-13\)
hay \(x=-\dfrac{13}{24}\)
BA
2
24 tháng 9 2016
a) (3x + 4)2 - (3x - 1).(3x + 1) = 49
=> (3x + 4).3x + (3x + 4).4 - (9x2 - 1) = 49
=> 9x2 + 12x + 12x + 16 - 9x2 + 1 = 49
=> 24x + 17 = 49
=> 24x = 49 - 17
=> 24x = 32
=> \(x=\frac{32}{24}=\frac{4}{3}\)
Vậy \(x=\frac{4}{3}\)
b) (2x + 1)2 - (x - 1)2 = 0
=> (2x + 1 - x + 1).(2x + 1 + x - 1) = 0
=> (x + 2).3x = 0
=> (x + 2).x = 0
\(\Rightarrow\left[\begin{array}{nghiempt}x+2=0\\x=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=-2\\x=0\end{array}\right.\)
Vậy \(\left[\begin{array}{nghiempt}x=-2\\x=0\end{array}\right.\)
NT
3
WH
10 tháng 5 2018
\(\left(3x+1\right)^2:\left(-\frac{1}{4}\right)=-\frac{49}{4}\)
\(\Leftrightarrow\left(3x+1\right)^2=-\frac{49}{4}\cdot\left(-\frac{1}{4}\right)\)
\(\Leftrightarrow\left(3x+1\right)^2=\frac{49}{16}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(3x+1\right)^2=\left(\frac{7}{4}\right)^2\\\left(3x+1\right)^2=\left(-\frac{7}{4}\right)^2\end{cases}\Leftrightarrow\orbr{\begin{cases}3x+1=\frac{7}{4}\\3x+1=-\frac{7}{4}\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=\frac{3}{4}\\3x=-\frac{11}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{11}{12}\end{cases}}}\)
VD
3
12 tháng 8 2021
( 3x + 4 )2 - ( 3x - 1 )( 3x + 1 ) = 49
<=> 9x2 + 24x + 16 - ( 9x2 - 1 ) - 49 = 0
<=> 9x2 + 24x - 33 - 9x2 + 1 = 0
<=> 24x - 32 = 0 <=> x = 4/3
AD
12 tháng 8 2021
\(\left(3x+4\right)^2-\left(3x-1\right)\left(3x+1\right)=49\)
\(\Leftrightarrow9x^2+24x+16-9x^2+1-49=0\)
\(\Leftrightarrow24x=32\)
\(\Leftrightarrow x=\frac{3}{2}\)
#H
DT
Dang Tung
CTVHS
15 tháng 10 2023
\(R=\left(\dfrac{3\sqrt{x}}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{3x-5\sqrt{x}}{4-x}\right):\left(\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}-1\right)\left(ĐK:x\ge0,x\ne4\right)\\ =\left(\dfrac{3\sqrt{x}}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{3x-5\sqrt{x}}{\sqrt{x}^2-2^2}\right):\dfrac{2\sqrt{x}-1-\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)+\sqrt{x}\left(\sqrt{x}+2\right)+3x-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}-2}{2\sqrt{x}-1-\sqrt{x}+2}\\ =\dfrac{3x-6\sqrt{x}+x+2\sqrt{x}+3x-5\sqrt{x}}{\sqrt{x}+2}.\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{7x-9\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)
Bạn xem lại đề nhé, rút gọn thường ra kết quả rất đẹp chứ không dài như kết quả này đâu ạ.
21 tháng 6 2018
(3x+4)\(^2\) - (3x-1)(3x+1)=49
=>\(9\text{x}^2+24x+16-9\text{x}^2+1\)\(=49\)
=>\(24\text{x}+17=49\)
=> 24x = 32
=> x = \(\dfrac{4}{3}\)
b) \(\left(3\text{x}-1\right)^2-\left(3\text{x}-2\right)^2=0 \)
\(=>9\text{x}^2-6\text{x}+1-9\text{x}^2+12\text{x}-4=0\)
\(=>6\text{x}-3=0\)
=> 6x = 3
=> x = \(\dfrac{1}{2}\)
c) \(\left(2\text{x}+1\right)^2-\left(x-1\right)^2=0\)
\(=>4\text{x}^2+4\text{x}+1-x^2+2\text{x}-1=0\)
=> \(3\text{x}^2+6\text{x}=0\)
=> \(3\text{x}\left(x+2\right)=0\)
=> 3x=0 hoặc x+2 = 0
+) 3x = 0 => x =0
+) x+2 = 0 => x = -2
YN
5 tháng 10 2021
a) \(\left(3x-5\right)\left(5-3x\right)+9\left(x+1\right)^2=30\)
\(\Rightarrow15x-9x^2-25+15x+9\left(x^2+2x+1\right)-30=0\)
\(\Rightarrow30x-9x^2-25+9x^2+18x+9-30=0\)
\(\Rightarrow48x-46=0\)
\(\Rightarrow x=\frac{23}{24}\)
b) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Rightarrow\left(x^2+8x+16\right)-\left(x^2-1\right)=16\)
\(\Rightarrow x^2+8x+16-x^2+1=16\)
\(\Rightarrow8x+17=16\)
\(\Rightarrow8x=-1\)
\(\Rightarrow x=\frac{-1}{8}\)
YN
5 tháng 10 2021
c) \(\left(y-2\right)^3-\left(y-3\right)\left(y^2+3y+9\right)+6\left(y+1\right)^2=49\)
\(\Rightarrow\left(y-2\right)^3-\left(y^3-3^3\right)+6\left(y^2+2y+1\right)=49\)
\(\Rightarrow y^3-6y^2+12y-8-y^3+27+6y^2+12y+6=49\)
\(\Rightarrow\left(y^3-y^3\right)+\left(-6y^2+6y^2\right)+\left(12y+12y\right)+\left(-8+27+6\right)=49\)
\(\Rightarrow24y+25=49\)
\(\Rightarrow24y=24\)
\(\Rightarrow y=1\)
d) \(\left(y+3\right)^3-\left(y+1\right)^3=56\)
\(\Rightarrow\left(y+3-y-1\right)[\left(y+3\right)^2+\left(y+3\right)\left(y+1\right)+\left(y+1\right)^2]=56\)
\(\Rightarrow2\left(y^2+6y+9+y^2+4y+3+y^2+2y+1\right)=56\)
\(\Rightarrow3y^2+12y+13=28\)
\(\Rightarrow\left(3y^2+15y\right)-\left(3y+15\right)=0\)
\(\Rightarrow3y\left(y+5\right)-3\left(y+5\right)=0\)
\(\Rightarrow3\left(y-1\right)\left(y+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
\(\left(3x+1\right)^2:\left(\frac{-1}{4}\right)=\frac{-49}{4}\)
\(\left(3x+1\right)^2=\frac{-49}{4}.\left(\frac{-1}{4}\right)\)
\(\left(3x+1\right)^2=\frac{49}{16}\)
\(\left(3x+1\right)^2=\frac{7}{4}^2\)
\(\Rightarrow3x+1=\frac{7}{4}\)
\(\Rightarrow3x=\frac{7}{4}-1\)
\(\Rightarrow3x=\frac{3}{4}\)
\(\Rightarrow x=\frac{3}{4}:3\)
\(\Rightarrow x=\frac{1}{4}\)
\(\left(3x+1\right)^2:\frac{-1}{4}=\frac{-49}{4}\)
\(\left(3x+1\right)^2=\frac{-49}{4}.\frac{-1}{4}\)
\(\left(3x+1\right)^2=3\frac{1}{6}\)
\(3x+1=1\frac{3}{4}\)
\(3x=1\frac{3}{4}+1\)
\(3x=2\frac{3}{4}\)
\(x=2\frac{3}{4}:3\)
\(x=\frac{11}{12}\)