Tìm x:
a, (x – 284) × 4 = 2864
b, (x : 7) + 286 = 3684
c, 78564 – x = 286
d, x + 2684 = 3895
78564 – y x 2 = 286
78564 - y x 2 = 286 y x 2 = 78278 y = 39139 Vậy y = 39139
Tìm x:
a) (x + 5)2 - x(x - 4) = 55
b) x(x - 7) - 3x + 21 = 0
a: \(\Leftrightarrow x^2+10x+25-x^2+4x=55\)
=>14x=30
hay x=15/7
b: \(\Leftrightarrow\left(x-7\right)\left(x-3\right)=0\)
hay \(x\in\left\{7;3\right\}\)
Tìm x:
a)x.(x+7)-(x-2).(x+3)=0
b)(x+2)2-(x2-4)=0
a: \(x\left(x+7\right)-\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow x^2+7x-x^2-x+6=0\)
hay x=-1
b: Ta có: \(\left(x+2\right)^2-\left(x^2-4\right)=0\)
\(\Leftrightarrow x+2=0\)
hay x=-2
b. (x + 2)2 - x2 + 4 = 0
<=> (x + 2 - x)(x + 2 + x) + 4 = 0
<=> 2(2 + 2x) + 4 = 0
<=> 4(1 + x) + 4 = 0
<=> 4(1 + x) = -4
<=> 1 + x = -1
<=> x = -1 - 1
<=> x = -2
\(a,\) \(x\left(x+7\right)-\left(x-2\right)\left(x+3\right)\)
\(\Leftrightarrow x^2+7x-x^2-3x+2x+6\\ \Leftrightarrow6x=0\\ \Leftrightarrow x=0\)
\(Vậy...\)
\(b,\) \(\left(x+2\right)^2-\left(x^2-4\right)=0\)
\(\Leftrightarrow x^2+4x+4-x^2+4=0\\ \Leftrightarrow4x+8=0\\ \Leftrightarrow x=-2\)
Tìm x:
a)x+3/4=5/3
b)x-2/3=7/2
a) x + \(\dfrac{3}{4}\) = \(\dfrac{5}{3}\)
x = \(\dfrac{5}{3}\) - \(\dfrac{3}{4}\)
x = \(\dfrac{20}{12}\) - \(\dfrac{9}{12}\)
x = \(\dfrac{11}{12}\)
b) x - \(\dfrac{2}{3}\) = \(\dfrac{7}{2}\)
x = \(\dfrac{7}{2}\) + \(\dfrac{2}{3}\)
x = \(\dfrac{21}{6}\) + \(\dfrac{4}{6}\)
x = \(\dfrac{25}{6}\)
Tìm x:
a) (x + 2) (x - 4) - x2 = 36
b) (x - 2)(4x + 1) - 4x(x + 7) = 1
c) x(x - 10) - x + 10 = 0
a: \(\Leftrightarrow x^2-2x-8-x^2=36\)
=>-2x=44
hay x=-22
b: \(\Leftrightarrow4x^2+x-8x-2-4x^2-27x=1\)
=>-34x=3
hay x=-3/34
c: =>(x-10)(x-1)=0
=>x=10 hoặc x=1
Tìm X:
a)(x-4)(x+4)=9
b)x2-4x+4-(5x-2)2=0
c)4x2+4+1-x2-10x-25=0
d)(x2+x+7)(x2+x-7)=(x2+x)2-7x
a)
⇔ \(x^2-16=9\)
⇔ \(x^2=25\)
⇔ \(x=\pm5\)
b)
⇔ \(x^2-4x+4-25x^2+20x-4=0\)
⇔ \(16x-24x^2=0\)
⇔ \(8x\left(2-3x\right)=0\)
⇒ \(\left[{}\begin{matrix}x=0\\2-3x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=\dfrac{2}{3}\)
c)
⇔ \(3x^2-10x-20=0\)
⇔ \(x^2-2.x.\dfrac{5}{3}+\dfrac{25}{9}-\dfrac{205}{9}=0\)
⇔ \(\left(x-\dfrac{5}{3}\right)^2=\dfrac{205}{9}\)
⇒ \(\left[{}\begin{matrix}x-\dfrac{5}{3}=\sqrt{\dfrac{205}{9}}\\x-\dfrac{5}{3}=-\sqrt{\dfrac{205}{9}}\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\\x=-\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\\\text{x}=-\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\end{matrix}\right.\)
Vậy...
d)
⇔ \(\left(x^2+x\right)^2-49=\left(x^2+x\right)^2-7x\)
⇔ 7x = 49
⇔ x=7
Vậy...
Tìm x:
a, 4.( x + 41 ) = 7
b, 4. ( x-3 ) = 7 mũ 2 - 1 mũ 10
a. 4.(x+41) = 7
x + 41 = 7 : 4 = 1,75
x = 1,75 - 41 = -39,25
b. 4.(x-3) = 72 - 110 = 49 - 1 = 48
x - 3 = 48 : 4 = 12
x = 12 + 3 = 15
Tìm x:
a, 4.( x + 41 ) = 400
b, 4. ( x-3 ) = 7 mũ 2 - 1 mũ 10
a) \(4\left(x+41\right)=400\)
\(\Rightarrow x+41=400:4\)
\(\Rightarrow x+41=100\)
\(\Rightarrow x=100-41\)
\(\Rightarrow x=59\)
a/ \(4.\left(x+41\right)=400\)
\(x+41=\dfrac{400}{4}=100\)
\(\Rightarrow x=59\)
b/ \(4.\left(x-3\right)=7^2-1^{10}\)
\(4.\left(x-3\right)=49-1\)
\(4.\left(x-3\right)=48\)
\(x-3=\dfrac{48}{3}=16\)
\(\Rightarrow x=19\)
#AEZn8
a ) 4 . ( x + 41 ) = 400
x + 41 = 400 : 4
x + 41 = 100
x = 100 - 41
x = 59
b ) 4 . ( x - 3 ) = 72- 110
4 . ( x - 3 ) = 49 - 1
4 . ( x - 3 ) = 48
x - 3 = 48 : 4
x - 3 = 12
x = 12 + 3
x = 15
Bài 2: Tìm X:
a) X x 12 + X x 7+ X = 100 c) (X- 60720) : 5= 318 + 642
b) X x 4 + X x 3 = 7497 :7 d) X : 4 + 11250 = 22850
Giải gấp cho em với ạ
\(a,\Leftrightarrow x\left(12+7+1\right)=100\)
\(\Leftrightarrow x.20=100\)
\(\Leftrightarrow x=50\)
\(b,\Leftrightarrow x\left(4+3\right)=1071\)
\(\Leftrightarrow7x=1071\)
\(\Leftrightarrow x=153\)
\(c,\Leftrightarrow x-60720=960\times5\)
\(\Leftrightarrow x-60720=4800\)
\(\Leftrightarrow x=4800+60720\)
\(\Leftrightarrow x=65520\)
\(d,\Leftrightarrow x:4=22850-11250\)
\(\Leftrightarrow x:4=11600\)
\(\Leftrightarrow x=46400\)