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\(\left(x+4\right)\left(x+6\right)\left(x-2\right)\left(x-12\right)=25x^2\)
\(\Leftrightarrow\left(x+3\right)\left(x+8\right)\left(x^2-15x+24\right)=0\)
\(x^4-8x^3+21x^2-24x+9=0\)
\(\Leftrightarrow\left(x^2-3x+3\right)\left(x^2-5x+3\right)=0\)
\(\Leftrightarrow\left(x-\frac{5+\sqrt{13}}{2}\right)\left(x-\frac{5-\sqrt{13}}{2}\right)=0\) (vì \(x^2-3x+3=\left(x-\frac{3}{2}\right)^2+0,75>0\))
\(\Rightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{13}}{2}\\x=\frac{5-\sqrt{13}}{2}\end{cases}}\)
a: =>(x^2+4x-5)(x^2+4x-21)=297
=>(x^2+4x)^2-26(x^2+4x)+105-297=0
=>x^2+4x=32 hoặc x^2+4x=-6(loại)
=>x^2+4x-32=0
=>(x+8)(x-4)=0
=>x=4 hoặc x=-8
b: =>(x^2-x-3)(x^2+x-4)=0
hay \(x\in\left\{\dfrac{1+\sqrt{13}}{2};\dfrac{1-\sqrt{13}}{2};\dfrac{-1+\sqrt{17}}{2};\dfrac{-1-\sqrt{17}}{2}\right\}\)
c: =>(x-1)(x+2)(x^2-6x-2)=0
hay \(x\in\left\{1;-2;3+\sqrt{11};3-\sqrt{11}\right\}\)
a)\(\left\{{}\begin{matrix}8x+2y=4\\8x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\4x+1=2\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}y=1\\x=\frac{1}{4}\end{matrix}\right.\)b)
\(\left\{{}\begin{matrix}12x-8y=44\\12x-15y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=35\\4x-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\4x-5.5=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=7\end{matrix}\right.\)c)\(\left\{{}\begin{matrix}9x=-18\\4x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\4.\left(-2\right)+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)
(1) \(\Leftrightarrow\left(x+1\right)\left(\sqrt{16x+17}-x+\dfrac{23}{8}\right)=0\)
cái này đâu ra z ???
nguyen van tuan: hì, xin lỗi, làm hơi tắt ^^!
\(\left(1\right)\Leftrightarrow\left(x+1\right)\sqrt{16x+17}=\left(x+1\right)\left(x-\dfrac{23}{8}\right)\Leftrightarrow\left(x+1\right)\sqrt{16x+17}-\left(x+1\right)\left(x-\dfrac{23}{8}\right)=0\Leftrightarrow\left(x+1\right)\left(\sqrt{16x+17}-x+\dfrac{23}{8}\right)=0\)
Lời giải:
Ta có:
\((x+3)(x+12)(x-4)(x-16)+20x^2=0\)
\(\Leftrightarrow [(x+3)(x-16)][(x+12)(x-4)]+20x^2=0\)
\(\Leftrightarrow (x^2-13x-48)(x^2+8x-48)+20x^2=0\)
Đặt \(x^2-12x-48=a\). PT trở thành:
\((a-x)(a+20x)+20x^2=0\)
\(\Leftrightarrow a^2+19ax-20x^2+20x^2=0\Leftrightarrow a^2+19ax=0\)
\(\Leftrightarrow a(a+19x)=0\)
\(\Leftrightarrow (x^2-12x-48)(x^2+7x-48)=0\)
\(\Leftrightarrow \left[\begin{matrix} x^2-12x-48=0\\ x^2+7x-48=0\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x=6\pm 2\sqrt{21}\\ x=\frac{-7\pm \sqrt{241}}{2}\end{matrix}\right.\)
Vậy......
i)
\(x^2-x^2\sqrt{2}-2x-2\sqrt{2}x+1+3\sqrt{2}=0\)
\(\left(x-1\right)^2+\sqrt{2}\left(x^2-2x+3\right)=0\)
\(\left(x-1\right)^2+\sqrt{2}\left(x-1\right)^2+2\sqrt{2}=0\)
\(\left(x-1\right)^2+\sqrt{2}\left(x-1\right)^2=-2\sqrt{2}\)
=> Phương trình vô nghiệm
ii)
Đặt: \(6x^2-7x=a\)
Ta có: \(a^2-2a-3=0\)
\(\left(a-3\right)\left(a+1\right)=0\)
\(\left(6x^2-7x-3\right)\left(6x^2-7x+1\right)=0\)
\(x=\frac{3}{2};-\frac{1}{3};1;\frac{1}{6}\)
Phương trình vô nghiệm
ii)
Đặt: $6x^2-7x=a$6x2−7x=a
Ta có: $a^2-2a-3=0$a2−2a−3=0
$\left(a-3\right)\left(a+1\right)=0$(a−3)(a+1)=0
$\left(6x^2-7x-3\right)\left(6x^2-7x+1\right)=0$(6x2−7x−3)(6x2−7x+1)=0
$
Sửa đề \(\left(8x-11\right)^3+\left(7x-12\right)^3+\left(23-15x\right)^3=0\)
Đặt \(8x-11=a\)
\(7x-12=b\)
\(23-15x=c\)
=> a+b+c=8x-11+7x-12+23-15x=0
Có \(a^3+b^3+c^3-3abc\)
= \(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
=\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2-3ab\right)\)
=0 (do a+b+c=0)
=> \(a^3+b^3+c^3=3abc\)
<=> \(0=3\left(8x-11\right)\left(7x-12\right)\left(23-15x\right)\)
=> \(\left[{}\begin{matrix}x=\frac{11}{8}\\x=\frac{12}{7}\\x=\frac{23}{15}\end{matrix}\right.\)