4x–1/3x²y—7x–1/3x²y
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YN
1
25 tháng 7 2021
Câu 5
Thay x = -2 vào pt y = -3x^2 + 7x - 5
y = -12 - 14 - 5 = -26 - 5 = - 31
Câu 6 Thay x = 3 vào y = ( 3x - 1)^2
y = ( 9 - 1 )^2 = 64
Câu 7 : Thay x = 1/2 y = \(\sqrt{x^2+4x}\)
= \(\sqrt{\dfrac{1}{4}+2}=\sqrt{\dfrac{9}{4}}=\dfrac{3}{2}\)
13 tháng 2 2022
1.
a.\(\Leftrightarrow7x-5x=3+12\)
\(\Leftrightarrow2x=15\Leftrightarrow x=\dfrac{15}{2}\)
b.\(\Leftrightarrow6x-10-7x-7=2\)
\(\Leftrightarrow x=-19\)
c.\(\Leftrightarrow1-3x=4x-3\)
\(\Leftrightarrow7x=2\Leftrightarrow x=\dfrac{2}{7}\)
d.\(\Leftrightarrow8x^2-4x+12x-6-8x^2-8x-2=12\)
\(\Leftrightarrow-2=12\left(voli\right)\)
H9
HT.Phong (9A5)
CTVHS
28 tháng 9 2023
Mình làm 5 câu thôi nhé !:
1) \(7^x\cdot49=7^{90}\)
\(\Rightarrow7^x\cdot7^2=7^{90}\)
\(\Rightarrow7^{x+2}=7^{90}\)
\(\Rightarrow x=90-2\)
\(\Rightarrow x=88\)
2) \(2^x\cdot4=128\)
\(\Rightarrow2^x\cdot2^2=2^7\)
\(\Rightarrow2^{x+2}=2^7\)
\(\Rightarrow x=7-2\)
\(\Rightarrow x=5\)
3) \(3^x-1=2^4\cdot5\)
\(\Rightarrow3^x=80+1\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
4) \(3^x+15=42\)
\(\Rightarrow3^x=42-15\)
\(\Rightarrow3^x=27\)
\(\Rightarrow3^x=3^3\)
\(\Rightarrow x=3\)
5) \(4\cdot2^x-3=125\)
\(\Rightarrow2^2\cdot2^x=128\)
\(\Rightarrow2^{x+2}=2^7\)
\(\Rightarrow x=7-2\)
\(\Rightarrow x=5\)
D
datcoder
CTVVIP
28 tháng 9 2023
6)
\(5^x=5^{15}:5^3\\ \Leftrightarrow5^x=5^{15-3}\\ \Leftrightarrow5^x=5^{12}\\ \Leftrightarrow x=12\)
7)
\(4^x=4^{15}:16\\ \Leftrightarrow4^x=4^{15}:4^2\\ \Leftrightarrow4^x=4^{15-2}\\ 4^x=4^{13}\\ \Leftrightarrow x=13\)
8)
\(7^x=7^{20}:7^{10}\\ \Leftrightarrow7^x=7^{20-10}\\ \Leftrightarrow7^x=7^{10}\\ \Leftrightarrow x=10\)
9)
\(11^x=11^{11}:11\\ \Leftrightarrow11^x=11^{11-1}\\ \Leftrightarrow11^x=11^{10}\\ \Leftrightarrow x=10\)
10)
\(3^{15}.3^x=3^{30}\\ \Leftrightarrow3^x=3^{30}:3^{15}\\ 3^x=3^{30-15}\\ \Leftrightarrow3^x=3^{15}\\ \Leftrightarrow x=15\)
C
2 tháng 12 2021
Tham Khảo:
https://olm.vn/hoi-dap/detail/264041645597.html
Sai thì hong bít j đâu ;-;
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 12 2021
\(7x^3+y^3+3xy\left(x-y\right)-12x^2+6x=1\)
\(\Leftrightarrow\left(8x^3-12x^2+6x-1\right)-\left(x^3-3x^2y+3xy^2-y^3\right)=0\)
\(\Leftrightarrow\left(2x-1\right)^3-\left(x-y\right)^3=0\)
\(\Leftrightarrow2x-1=x-y\)
\(\Leftrightarrow y=1-x\)
Thế xuống dưới:
\(\sqrt[3]{3x+2}+\sqrt{x+2}=4\)
\(\Leftrightarrow\sqrt[3]{3x+2}-2+\sqrt{x+2}-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{\sqrt[3]{\left(3x+2\right)^2}+2\sqrt[3]{3x+2}+4}+\dfrac{1}{\sqrt{x+2}+2}\right)=0\)
25 tháng 1 2021
a) Ta có: \(x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy: S={-5;2}
b) Ta có: \(3x^2-7x+1=0\)
\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)=0\)
mà 3>0
nên \(x^2-\dfrac{7}{3}x+\dfrac{1}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{37}{36}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{6}=\dfrac{\sqrt{37}}{6}\\x-\dfrac{7}{6}=-\dfrac{\sqrt{37}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{37}+7}{6}\\x=\dfrac{-\sqrt{37}+7}{6}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{\sqrt{37}+7}{6};\dfrac{-\sqrt{37}+7}{6}\right\}\)
c) Ta có: \(3x^2-7x+8=0\)
\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{8}{3}\right)=0\)
mà 3>0
nên \(x^2-\dfrac{7}{3}x+\dfrac{8}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{47}{36}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=-\dfrac{47}{36}\)(vô lý)
Vậy: \(x\in\varnothing\)
\(\frac{4x-1}{3x^2y}-\frac{7x-1}{3x^2y}\)
\(=\frac{\left[\left(4x-1\right)-\left(7x-1\right)\right]}{3x^2y}\)
\(=\frac{\left(4x-1-7x+1\right)}{3x^2y}\)
\(=\frac{\left(-3x\right)}{3x^2y}\)
\(\frac{-1}{xy}\)