a)|x-5|=x-5
b)|x-6|=6-x
GIÚP MÌNH NHÉ
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\(x:\left[\dfrac{8}{5}\cdot\left(\dfrac{2}{3}\right)^2-\dfrac{2}{5}\right]=\dfrac{15}{7}+\dfrac{6}{5}\left[\left(2\dfrac{1}{7}\right)^2-\dfrac{50}{49}\right]\)
\(\Leftrightarrow x:\left[\dfrac{32}{45}-\dfrac{18}{45}\right]=\dfrac{15}{7}+\dfrac{6}{5}\cdot\left(\dfrac{225}{49}-\dfrac{50}{49}\right)\)
\(\Leftrightarrow x:\dfrac{14}{45}=\dfrac{15}{7}+\dfrac{6}{5}\cdot\dfrac{25}{7}\)
\(\Leftrightarrow x:\dfrac{14}{45}=\dfrac{45}{7}\)
\(\Leftrightarrow x=2\)
Đề trước đó:
(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x
<=>x^2+x-7x-7-x^2+6x-9=9x^2-25-9x^2-6x-1+x^2-4x+4-x
<=>x^2-11x-6=0
<=>x^2-2x. 11/2 + 121/4-145/4=0
<=>(x-11/2)^2=145/4
<=>|x-11/2|=căn(145)/2
<=>x=[11+-căn(145)]/2
a)
\(\Rightarrow\dfrac{47}{10}.x=\dfrac{1}{5}+\dfrac{2}{3}\)
\(\Rightarrow\dfrac{47}{10}.x=\dfrac{3}{15}+\dfrac{10}{15}\)
\(\Rightarrow\dfrac{47}{10}.x=\dfrac{13}{15}\)
\(\Rightarrow x=\dfrac{13}{15}:\dfrac{47}{10}\)
\(\Rightarrow x=\dfrac{13}{10}.\dfrac{10}{47}\)
\(\Rightarrow x=\dfrac{13}{47}\)
b)
\(\Rightarrow\dfrac{5}{7}:x=\dfrac{1}{6}-\dfrac{4}{5}\)
\(\Rightarrow\dfrac{5}{7}:x=\dfrac{5}{30}-\dfrac{24}{30}\)
\(\Rightarrow\dfrac{5}{7}:x=-\dfrac{19}{30}\)
\(\Rightarrow x=\dfrac{5}{7}:\left(-\dfrac{19}{30}\right)\)
\(\Rightarrow x=\dfrac{5}{7}.\left(-\dfrac{30}{19}\right)\)
\(\Rightarrow x=-\dfrac{150}{133}\)
ĐK:\(x\ge0\)
\(\left(x^2-1\right)\sqrt{x}=0\Leftrightarrow\left(x-1\right)\left(x+1\right)\sqrt{x}=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\\sqrt{x}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(ktm\right)\\x=0\left(tm\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)
Ủa lớp 7 sao học căn r nè
a, x >=5
b,x=6
LÊ HOÀNG BẢO NGÂN:
a) \(\left|x-5\right|=x-5\)
\(\Leftrightarrow x\ge5\)
b) \(\left|x-6\right|=6-x\)
\(\Leftrightarrow x=6\)
(^_^). Ủng hộ mk nha các bạn!!!