câu
x/-35=6/14
x=?
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[(14x+28)*3+55}:5=35
(14x+28)*3+55=35*5
(14x+28)*3+55=175
(14x+28)*3=175-55
(14x+28)*3=120
14x+28=120:3
14x+28=40
14x=40-28
14x=12
x=12:14
x=6/7
a) 6x - 35 + 5x - 21 = (6x + 5x) - (35 + 21) = 11x - 56 = 17x - 20 => 11x = 17x - 20 + 56 = 17x + 36
=> 36 = 11x - 17x = -6x => x = 36 : (-6) = -6
b) 5x + 35 - 4x - 27 = (5x - 4x) + (35 - 27) = x + 8 = 14x + 34 => x = 14x + 34 - 8 = 14x + 26
=> 26 = x - 14x = -13x => x = 26 : (-13) = -2
a) \(6x-35+5x-21=17x-20\)
\(\Rightarrow6x-25+5x-21+20=17x\)
\(\left(6x+5x\right)+\left(20-25-21\right)=17x\)
\(11x+\left(-36\right)=17x\)
\(17x-11x=-36\)
\(6x=-36\)
\(x=\frac{-36}{6}=-6\)
Vậy \(x=-6\)
b) \(5x+35-4x-27=14x+34\)
\(\Rightarrow5x+35-4x-27-34=14x\)
\(\left(5x-4x\right)+\left(35-27-34\right)=14x\)
\(x+\left(-26\right)=14x\)
\(14x-x=-26\)
\(13x=-26\)
\(x=-\frac{26}{13}=-2\)
Vậy \(x=-2\)
`@` `\text {Ans}`
`\downarrow`
`a,`
`x - 280 \div 35 = 554`
`=> x - 8 = 554`
`=> x = 554 + 8`
`=> x = 562`
Vậy, `x = 562`
`b,`
`x + (5 \div 3) = 500`
`=> x + 5/3 = 500`
`=> x = 500 - 5/3`
`=> x = 1495/3`
Vậy, `x = 1495/3`
`c,`
`390 \div (5x - 5) = 39`
`=> 5x - 5 = 390 \div 39`
`=> 5x - 5 = 10`
`=> 5x = 10 + 5`
`=> 5x = 15`
`=> x = 15 \div 5`
`=> x = 3`
Vậy, `x = 3`
`d,`
`34x - 14x = 200`
`=> (34 - 14)x = 200`
`=> 20x = 200`
`=> x = 200 \div 20`
`=> x = 10`
Vậy, `x = 10.`
`@` `\text {Kaizuu lv uuu}`
a) \(x-280:35=554\)
\(x-8=554\)
\(x=562\)
b) \(x+\left(5:3\right)=500\)
\(x+\dfrac{5}{3}=500\)
\(x=500-\dfrac{5}{3}=\dfrac{1500}{3}-\dfrac{5}{3}=\dfrac{1495}{3}\)
c) \(390:\left(5x-5\right)=39\)
\(5x-5=\dfrac{39}{390}=\dfrac{1}{10}\)
\(5x=5+\dfrac{1}{10}\)
\(5x=\dfrac{51}{10}\)
\(x=\dfrac{51}{10}.\dfrac{1}{5}=\dfrac{51}{50}\)
d) \(34x-14x=200\)
\(20x=200\)
\(x=200:20=10\)
7\(x^2\)+\(3y^2+z^2-14x+2z-18y+35=0\)
\(\Leftrightarrow\left(7x^2-14x+7\right)+\left(3y^2-18y+27\right)+\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow7\left(x-1\right)^2+3\left(y-3\right)^2+\left(z+1\right)^2=0\)
mà \(\left(x-1\right)^2\ge0\forall x\);\(\left(y-3\right)^2\ge0\forall y\);\(\left(z+1\right)^2\ge0\forall z\)\(\Rightarrow\)phương trình có nghiệm khi đồng thời x-1=0;
y-3=0;z+1=0hay x=1;y=3;z=-1
x, y là số thực, x < hoặc bằng 2, x + y >hoặc = 2. Tìm GTNN của A = 14x2+ 9y2+ 22xy - 42x - 34y + 35
128 - 3 ( x + 4 ) = 23
3 ( x + 4 ) = 128 - 23
3 ( x + 4 ) = 105
x + 4 = 105 : 5
x + 4 = 35
x = 35 - 4
x = 31
128 - 3 ( x + 4 ) = 23
3 ( x + 4 ) = 128 - 23
3 ( x + 4 ) = 105
x + 4 = 105 : 5
x + 4 = 35
x = 35 - 4
x = 31
k mik nha
a/ ĐKXĐ: \(x\ge1\)
Khi \(x\ge1\) ta thấy \(\left\{{}\begin{matrix}VT>0\\VP=1-x\le0\end{matrix}\right.\) nên pt vô nghiệm
b/ \(x\ge1\)
\(\sqrt{\sqrt{x-1}\left(x-2\sqrt{x-1}\right)}+\sqrt{\sqrt{x-1}\left(x+3-4\sqrt{x-1}\right)}=\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{\sqrt{x-1}\left(\sqrt{x-1}-1\right)^2}+\sqrt{\sqrt{x-1}\left(\sqrt{x-1}-2\right)^2}=\sqrt{x-1}\)
Đặt \(\sqrt{x-1}=a\ge0\) ta được:
\(\sqrt{a\left(a-1\right)^2}+\sqrt{a\left(a-2\right)^2}=a\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\Rightarrow x=1\\\sqrt{\left(a-1\right)^2}+\sqrt{\left(a-2\right)^2}=\sqrt{a}\left(1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left|a-1\right|+\left|a-2\right|=\sqrt{a}\)
- Với \(a\ge2\) ta được: \(2a-3=\sqrt{a}\Leftrightarrow2a-\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}\sqrt{a}=-1\left(l\right)\\\sqrt{a}=\frac{3}{2}\end{matrix}\right.\)
\(\Rightarrow a=\frac{9}{4}\Rightarrow\sqrt{x-1}=\frac{9}{4}\Rightarrow...\)
- Với \(0\le a\le1\) ta được:
\(1-a+2-a=\sqrt{a}\Leftrightarrow2a+\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x-1}=1\Rightarrow...\)
- Với \(1< a< 2\Rightarrow a-1+2-a=\sqrt{a}\Leftrightarrow a=1\left(l\right)\)
c/ ĐKXĐ: \(x\ge\frac{49}{14}\)
\(\Leftrightarrow\sqrt{14x-49+14\sqrt{14x-49}+49}+\sqrt{14x-49-14\sqrt{14x-49}+49}=14\)
\(\Leftrightarrow\sqrt{\left(\sqrt{14x-49}+7\right)^2}+\sqrt{\left(\sqrt{14x-49}-7\right)^2}=14\)
\(\Leftrightarrow\left|\sqrt{14x-49}+7\right|+\left|7-\sqrt{14x-49}\right|=14\)
Mà \(VT\ge\left|\sqrt{14x-49}+7+7-\sqrt{14x-49}\right|=14\)
Nên dấu "=" xảy ra khi và chỉ khi:
\(7-\sqrt{14x-49}\ge0\)
\(\Leftrightarrow14x-49\le49\Leftrightarrow x\le7\)
Vậy nghiệm của pt là \(\frac{49}{14}\le x\le7\)
x=248/7
theo đề bài suy ra x=6*(-35)/14=-15
vậy x=-15