Bài 1 : Tìm x
a) ( 2x x +28 ) + 55 = 175 - b)41-(2x x -5)=22 - c) ( 5x x +16):3 =7
Bài 2.Tính
1) 2+37+135+58+63 - 2) 25x17x8x4x125 - 3) 36x23+62x23+46 - 4) 42x167-42x67
Ai giải giùm em với ạ em cảm ơn^^
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Những câu hỏi liên quan
30 tháng 6 2021
a) Ta có: \(8x\left(2x-3\right)-4x\left(4x+3\right)=72\)
\(\Leftrightarrow16x^2-24x-16x^2-12x=72\)
\(\Leftrightarrow-36x=72\)
hay x=-2
b) Ta có: \(\left(x+2\right)\left(x+4\right)-x\left(x+2\right)=104\)
\(\Leftrightarrow x^2+6x+8-x^2-2x=104\)
\(\Leftrightarrow4x=96\)
hay x=24
c) Ta có: \(\left(x-1\right)\left(x+4\right)-x\left(x-1\right)=308\)
\(\Leftrightarrow x^2+3x-4-x^2+x=308\)
\(\Leftrightarrow4x=312\)
hay x=78
d) Ta có: \(15x\left(2x-3\right)-\left(5x+2\right)\left(6x-5\right)=-22\)
\(\Leftrightarrow30x^2-45x-30x^2+25x-12x+10=-22\)
\(\Leftrightarrow-32x=-32\)
hay x=1
DT
13
VN
Võ Ngọc Phương
CTVHS
VIP
26 tháng 8 2023
a) \(\dfrac{1}{4}+\dfrac{3}{4}:x=-2\)
\(\dfrac{3}{4}:x=-2-\dfrac{1}{4}=\dfrac{-8}{4}-\dfrac{1}{4}\)
\(\dfrac{3}{4}:x=\dfrac{-9}{4}\)
\(x=\dfrac{3}{4}:\dfrac{-9}{4}=\dfrac{3}{4}.\dfrac{-4}{9}\)
\(x=\dfrac{-1}{3}\)
b) \(\dfrac{3}{4}+2.\left(2x-\dfrac{2}{3}\right)=-2\)
\(2.\left(2x-\dfrac{2}{3}\right)=-2-\dfrac{3}{4}=\dfrac{-8}{4}-\dfrac{3}{4}\)
\(2.\left(2x-\dfrac{2}{3}\right)=\dfrac{-11}{4}\)
\(2x-\dfrac{2}{3}=\dfrac{-11}{4}:2=\dfrac{-11}{4}.\dfrac{1}{2}\)
\(2x-\dfrac{2}{3}=\dfrac{-11}{8}\)
\(2x=\dfrac{-11}{8}+\dfrac{2}{3}=\dfrac{-33}{24}+\dfrac{16}{24}\)
\(2x=\dfrac{-17}{24}\)
\(x=\dfrac{-17}{24}:2=\dfrac{-17}{24}.\dfrac{1}{2}\)
\(x=\dfrac{-17}{48}\)
c) \(\left(\dfrac{1}{2}+5x\right).\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}+5x=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{-1}{2}\\2x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{10}\\x=\dfrac{3}{2}\end{matrix}\right.\)
26 tháng 8 2023
a, 1/4 + 3/4 : x = -2
3/4 : x = -2 - 1/4
3/4 : x = -9/4
x = 3/4 : -9/4
x = -1/3
T
22 tháng 10 2023
\(a,(x-2)^2-25=0\\\Leftrightarrow (x-2)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
\(---\)
\(b,4x(x-2)+x-2=0\\\Leftrightarrow4x(x-2)+(x-2)=0\\\Leftrightarrow(x-2)(4x+1)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{4}\end{matrix}\right.\)
\(---\)
\(c,4x(x-2)-x(3+4x)(?)\)
\(d,(2x-5)^2-3x(5-2x)=0\\\Leftrightarrow(2x-5)^2+3x(2x-5)=0\\\Leftrightarrow(2x-5)(2x-5+3x)=0\\\Leftrightarrow(2x-5)(5x-5)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\5x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=1\end{matrix}\right.\)
\(---\)
\(e,x^2-25-(x+5)=0(sửa.đề)\\\Leftrightarrow(x^2-5^2)-(x+5)=0\\\Leftrightarrow (x-5)(x+5)-(x+5)=0\\\Leftrightarrow(x+5)(x-5-1)=0\\\Leftrightarrow(x+5)(x-6)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
\(---\)
\(f,5x(x-3)-x+3=0\\\Leftrightarrow5x(x-3)-(x-3)=0\\\Leftrightarrow(x-3)(5x-1)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
\(Toru\)
9 tháng 10 2021
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
14 tháng 6 2016
a,
(x2-x+1)(x+1)-x3+3x=15
x3-x2+x+x2-x+1-x3+3x=15
x3-x3-x2+x2+x-x+3x+1=15
3x+1=15
3x=15-1
3x=14
x=14/3
b,
(x+3)(x-2)+3x=\(\frac{4}{x+\frac{3}{4}}\)
x2-2x+3x-6+3x=\(\frac{4}{x+\frac{3}{4}}\)
x2-2x+3x+3x-6=\(\frac{4}{x+\frac{3}{4}}\)
Tới đây hết biết , đề có gì sai sai sao ý !
c,
(x2-5)(x+2)+5x=2x2+17
x3+2x2-5x-10+5x=2x2+17
x3+2x2-5x+5x-10=2x2+17
x3+2x2-10=2x2+17
x3-10=17
x3=17+10
x3=27
\(\Rightarrow x=3\)(Vì : 33=27)
_k_ nhé bn
14 tháng 6 2016
Nhân ra thôi bạn, có hằng đẳng thức gì đâu !
a) \(\left(x^2-x+1\right)\left(x+1\right)-x^3+3x=15\)
\(\Leftrightarrow\left(x^2-x+1\right)\cdot x+x^2-x+1-x^3+3x=15\)
\(\Leftrightarrow x^3-x^2+x+x^2-x+1-x^3+3x=15\)
\(\Leftrightarrow1+3x=15\Leftrightarrow3x=14\Leftrightarrow x=\frac{14}{3}\)
b) \(\left(x+3\right)\left(x-2\right)+3x=4\cdot\left(x+\frac{3}{4}\right)\)
\(\Leftrightarrow x^2+3x-2x-6+3x=4x+3\)
\(\Leftrightarrow x^2+4x-6=4x+3\)
\(\Leftrightarrow x^2=9\Leftrightarrow\orbr{\begin{cases}x=-3\\x=3\end{cases}}\)
c) \(\left(x^2-5\right)\left(x+2\right)+5x=2x^2+17\)
\(\Leftrightarrow x^3-5x+2x^2-10+5x=2x^2+17\)
\(\Leftrightarrow x^3=27\Leftrightarrow x=3\)
CH
Cô Hoàng Huyền
Admin
VIP
14 tháng 6 2016
Chúng ta sẽ sử dụng hằng đẳng thức em nhé :)
a. \(x^3+1-x^3+3x=15\Leftrightarrow3x=14\Leftrightarrow x=\frac{14}{3}\)
b. \(x^2+x-6+3x=4x+3\Leftrightarrow x^2=9\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
c. \(x^3+2x^2-5x-10+5x=2x^2+17\Leftrightarrow x^3=27\Leftrightarrow x=3\)
14 tháng 6 2016
a)\(\left(x^2-x+1\right).\left(x+1\right)-x^3+3x=15\)
\(x^3+x^2-x^2-x+x+1-x^3+3x=15\)
\(1+3x=15\)
\(3x=15-1\)
\(3x=14\)
\(x=\frac{14}{3}\)
14 tháng 6 2016
b) \(\left(x+3\right).\left(x-2\right)+3x=4\left(x+\frac{3}{4}\right)\)
\(x^2-2x+3x-6+3x=4x+3\)
\(x^2-2x+3x+3x-4x=6+3\)
\(x^2=9\)
\(x^2=3^2\) hoặc \(x^2=\left(-3\right)^2\)
vậy x=3 hoặc x=-3
V
1
AH
Akai Haruma
Giáo viên
18 tháng 6 2021
Nếu bạn thiếu số 2 bên cạnh $\sqrt{2x^2+5x+3}$ thì có thể tham khảo lời giải tại đây:
https://hoc24.vn/cau-hoi/tim-x-sao-cho-sqrt2x3sqrtx13x2sqrt2x25x3-16.235781793134
