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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) \(\left(y+3\right)^3-\left(y+1\right)^3=56\)
\(\Rightarrow\left(y+3-y-1\right)[\left(y+3\right)^2+\left(y+3\right)\left(y+1\right)+\left(y+1\right)^2]=56\)
\(\Rightarrow2\left(y^2+6y+9+y^2+4y+3+y^2+2y+1\right)=56\)
\(\Rightarrow3y^2+12y+13=28\)
\(\Rightarrow\left(3y^2+15y\right)-\left(3y+15\right)=0\)
\(\Rightarrow3y\left(y+5\right)-3\left(y+5\right)=0\)
\(\Rightarrow3\left(y-1\right)\left(y+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
28 tháng 8 2023
a) \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{x^2-y^2}{4-9}=\dfrac{-16}{-5}=\dfrac{16}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=4.\dfrac{16}{5}\\y^2=9.\dfrac{16}{5}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\pm\left(2.\dfrac{4}{\sqrt[]{5}}\right)=\pm\dfrac{8\sqrt[]{5}}{5}\\y=\pm\left(3.\dfrac{4}{\sqrt[]{5}}\right)=\pm\dfrac{12\sqrt[]{5}}{5}\end{matrix}\right.\)
\(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow z=\dfrac{5}{4}y=\dfrac{5}{4}.\left(\pm\dfrac{12\sqrt[]{5}}{5}\right)=\pm3\sqrt[]{5}\)
b) \(\left|2x+3\right|=x+2\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=x+2\\2x+3=-x-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\3x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\3x=-\dfrac{5}{3}\end{matrix}\right.\)
28 tháng 8 2023
Đính chính
Dòng cuối \(3x=-\dfrac{5}{3}\rightarrow x=-\dfrac{5}{3}\)
LN
1 tháng 11 2015
a\(\left(x-3\right)^2-\left(x+2\right)^2-5\left(\frac{1}{5}-7\right)=-30\)
=>(x-3-x-2)(x-3+x+2)-x+35=-30
=>-5(2x-1)-x+35=-30
=>-10x+5-x+35=-30
=>-11x+40=-30
=>-11x=-70 =>x=70/11
d)\(\left(x+3\right)^2-\left(x+5\right)\left(x-5\right)=2\)
\(=>\left(x+3\right)^2-x^2+25=2\)
\(=>\left(z+3-z\right)\left(z+3+z\right)+25=2\)
\(=>3\left(2z+3\right)+25-2=0\)
\(=>6z+9+23=0\)
\(=>6x+32=0=>6x=-32=>x=-\frac{16}{3}\)
e)\(3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\)
\(=>3\left(x^2+4x+4\right)+\left(4x^2-4x+1\right)-7\left(x^2-9\right)=36\)
\(=>3x^2+12x+12+4x^2-4x+1-7x^2+63\)
\(=>8x+76=36=>8x=36-76=>x=-40\div8=-5\)
g)\(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)
\(=>x^3-1-x\left(x^2-4\right)=5=>x^3-1-x^3+4x=5\)
\(=>4x-1=5=>4x=6=>x=\frac{3}{2}\)
AH
Akai Haruma
Giáo viên
4 tháng 11 2023
Bài 1:
$2x(x+3)+(2x+3)(5-x)=2$
$\Leftrightarrow 2x^2+6x+(10x-2x^2+15-3x)=2$
$\Leftrightarrow 2x^2+6x+7x-2x^2+15=2$
$\Leftrightarrow 13x+15=2$
$\Leftrightarrow 13x=2-15=-13$
$\Leftrightarrow x=-13:13=-1$
AH
Akai Haruma
Giáo viên
4 tháng 11 2023
Bài 2:
$x-y=4\Rightarrow x=y+4$. Thay vào $xy=5$ thì:
$(y+4)y=5$
$\Leftrightarrow y^2+4y-5=0$
$\Leftrightarrow (y-1)(y+5)=0$
$\Leftrightarrow y=1$ hoặc $y=-5$
Nếu $y=1$ thì $x=y+4=5$. Khi đó $x^3+y^3=5^3+1^3=126$
Nếu $y=-5$ thì $x=y+4=-1$. Khi đó: $x^3+y^3=(-1)^3+(-5)^3=-126$
H
3
25 tháng 8 2023
\(\dfrac{x}{3}=\dfrac{y}{4}\Leftrightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}\)
\(\dfrac{z}{5}=\dfrac{z^2}{25}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{x^2+y^2}{9+16}=\dfrac{x^2+y^2}{25}=\dfrac{225}{25}=9\)
\(\Rightarrow x=\sqrt{9\cdot9}=9\)
\(\Rightarrow y=\sqrt{9\cdot16}=12\)
\(\Rightarrow z=\sqrt{9\cdot25}=15\)
25 tháng 8 2023
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)
\(\Rightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=\dfrac{225}{25}=9\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=9.9=81\\y^2=16.9=144\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=9\\y=12\end{matrix}\right.\)
\(\Rightarrow z=\dfrac{9}{3}.5=15\)
Vậy \(\left\{{}\begin{matrix}x=9\\y=12\\z=15\end{matrix}\right.\) thỏa đề bài
Không có dấu '' = '' để tìm x nhé.
\([4.\left(x-y\right)^5+2.\left(x-y\right)^3-3.\left(x-y\right)^2];\left(y-x\right)^2\)
\(=[4.\left(x-y\right)^5+2.\left(x-y\right)^3-3.\left(x-y\right)^2]:\left(x-y\right)^2\)
\(=4.\left(x-y\right)^3+2.\left(x-y\right)-3\)
\(=4.\left(x^3-3x^2y+3xy^2-y^3\right)+2x-2y-3\)
\(=4x^3-12x^2y+12xy^2-y^3+2x-2y-3\)