-5x= 4+7
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Tìm x biết
(5x-1)2-(5x-4)(5x+4)=7
`(5x-1)^2-(5x-4)(5x+4)=7`
`\Leftrightarrow 25x^2-10x+1-25x^2+16-7=0`
`\Leftrightarrow -10x+10=0`
`\Leftrightarrow x=1`
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Tính
\(x^7-5x^6+5x^5-5x^4+5x^3-5x+5x-7\)
(5x-1)2 - (5x-4) (5x+4)=7
Tìm x
(5x-1)2 _ (5x-4)(5x+4)= 7
\(\left(5x-1\right)^2-\left(5x-4\right)\left(5x+4\right)=7^{ }\)
\(\Leftrightarrow25x^2-10x+1-25x^2+16=7\)
\(\Leftrightarrow17-10x=7\)
\(\Leftrightarrow10-10x=0\)
\(\Rightarrow x=1\)
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(5x-1)^2+(5x-4)×(5x+4)=7
Tìmx
Ta có: \(\left(5x-1\right)^2+\left(5x-4\right)\left(5x+4\right)=7\)
\(\Leftrightarrow25x^2-10x+1+25x^2-16-7=0\)
\(\Leftrightarrow25x^2-10x-22=0\)
\(\Leftrightarrow\left(5x\right)^2-2\cdot5x\cdot1+1-23=0\)
\(\Leftrightarrow\left(5x-1\right)^2=23\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=\sqrt{23}\\5x-1=-\sqrt{23}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\sqrt{23}+1\\5x=-\sqrt{23}+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{23}+1}{5}\\x=\frac{-\sqrt{23}+1}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{\sqrt{23}+1}{5};\frac{-\sqrt{23}+1}{5}\right\}\)
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(x^2-5x+9)^2 + 4(x^2-3x+7)^2 = 4(x^2-5x+9)(x^2-3x+7)
12 tháng 3 2022 lúc 16:58
thu gọn đa thức 2x^3+2x^5-5x^7-7x^2-11x^3+2, 5x^4-9+4, 2x^2+1, 5x^4+13x^8
Giải phương trình sau :
a,\(\dfrac{7-3x}{12}+\dfrac{5x+2}{7}=x+13\)
b,\(\dfrac{3\left(x+3\right)}{4}-\dfrac{1}{2}=\dfrac{5x+9}{7}-\dfrac{7x-9}{4}\)
c,\(\dfrac{2x+1}{3}-\dfrac{5x+2}{7}=x+3\)
d,\(\dfrac{2x-3}{3}-\dfrac{2x+3}{7}=\dfrac{4x+3}{5}-17\)
a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)
\(\Leftrightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow39x-84x=1092-73\)
=>-45x=1019
hay x=-1019/45
b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
=>21x+63-14=20x+36-49x+63
=>21x+49=-29x+99
=>50x=50
hay x=1
c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)
=>14x+7-15x-6-21x-63=0
=>-22x-64=0
hay x=-32/11
d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)
=>70x-105-30x-45=84x+63-1785
=>40x-150-84x+1722=0
=>-44x+1572=0
hay x=393/11
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a, msc 12.7=84
Chuyển vế về =0 rồi làm
b,msc 28
c,làm tương tự
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a, \(\Rightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow-45x=1019\Leftrightarrow x=-\dfrac{1019}{45}\)
b, \(\Rightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
\(\Leftrightarrow21x+63-14=20x+36-49x+63\)
\(\Leftrightarrow50x=50\Leftrightarrow x=1\)
c, \(\Rightarrow14x+7-15x-6=21x+63\Leftrightarrow-22x=62\Leftrightarrow x=-\dfrac{31}{11}\)
d, \(\Rightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-105.17\)
\(\Leftrightarrow70x-105-30x-45=84x+63-1785\)
\(\Leftrightarrow-44x=-1572\Leftrightarrow x=\dfrac{393}{11}\)
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Xem thêm câu trả lời
∣5x+7∣∣∣+∣∣∣5x−1∣∣∣3.23(2y+1)2020+4
Đọc tiếp
∣5x+7∣∣∣+∣∣∣5x−1∣∣∣=3.23(2y+1)2020+4
5x+7+5x-1=(5x+5x)+(7-1)=10x+6
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1).(4-3x)(10-5x)=0 2).(7-2x)(4+8x)=0 3).(9-7x)(11-3x)=0
4).(7-14x)(x-2)=0 5).(\(\dfrac{7}{8}\)-2x)(3x+\(\dfrac{1}{3}\))=0 6).3x-2x\(^2\)
7).5x+10x\(^2\)
1.
<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)
2.
<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
3.
<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)
4.
<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)
5.
<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)
6,7. ko đủ điều kiện tìm
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