tìm x
a)\(2^{x+1}=32\)
b) 3+3^x-2=810
c) \(\left(7^x\right)^2=7^{14}\)
d) \(\left(\left(-0.5\right)^x\right)=\frac{1}{32}\)
tìm x:
a,\(\left(x-\frac{7}{9}\right)^3=\left(\frac{2}{3}\right)^6\)
b,\(\left(\frac{x}{2}-5\right)^5=-32\)
c,\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
a,\(\left(x-\frac{7}{9}\right)^3=\left(\left(\frac{2}{3}\right)^2\right)^3\)
\(x-\frac{7}{9}=\frac{4}{9}\)
\(x=\frac{4}{9}+\frac{7}{9}\)
\(x=\frac{11}{9}\)
Vậy x=\(\frac{11}{9}\)
Tìm x biết:
a, \(\left(4,5-2.x\right).1\dfrac{4}{7}=\dfrac{11}{14}\)
b, \(\left(2,8.x-32\right):\dfrac{2}{3}=-90\)
Cảm ơn ạ!
`a)(4,5-2x)*1 4/7=11/14`
`=>(4,5-2x)*11/7=11/14`
`=>4,5-2x=1/2`
`=>2x=4,5-0,5=4`
`=>x=2`
Vậy `x=2`
`b)(2,8x-32):2/3=-90`
`=>2,8x-32=-90*2/3=-60`
`=>2,8x=-28`
`=>x=-10`
Vậy `x=-10`
Tìm x biết:
\(a,\left(x-\dfrac{3}{4}\right)+50\%=\dfrac{1}{6}\)
\(b,\dfrac{1}{2}x-\dfrac{5}{6}x=\dfrac{7}{2}\)
\(c,\left(4-x\right)\left(3x+5\right)=0\)
\(d,\dfrac{x}{16}=\dfrac{50}{32}\)
\(e,\left(2x-3\right)+\dfrac{3}{2}=-\dfrac{1}{4}\)
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
Tìm x, y biết:
a) \(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
b) (3-x) : 0,16 = -9 : (x-3)
c) \(\left(3x-\frac{4}{5}\right)^2+\left(2y+\frac{3}{7}\right)^2=0\)
d) \(\left(x+y-\frac{1}{2}\right)^2+\left(x-y+\frac{1}{6}\right)^2=0\)
e) x. \(\left(x-\frac{1}{4}\right)\)> 0
Tính
a) \(\frac{-1}{7}.\left(9\frac{1}{2}-8,75\right):\frac{2}{7}+0,625:1\frac{2}{3}\)
b)x-25%.x-1=-|-5|
c)\(2\frac{3}{4}.\left(0,4\right)1\frac{3}{5}.2,75+\left(-1,2\right):\frac{4}{11}\)
d) (2,8 x -32 ):\(\frac{2}{3}=-90\)
2. tìm x
a) \(\left(x-1\right)^3=8\)
b) \(7^{2x-6}=49\)
c) \(\left(2x-14\right)^7=128\)
d) \(x^4.x^5=5^3.5^6\)
e) \(\left[3.\left(x+2\right):7\right].4=120\)
a) \(\left(x-1\right)^3=8=2^3\)
\(x-1=2\)
\(x=2+1=3\)
b) \(7^{2x-6}=49=7^2\)
\(2x-6=2\)
\(2x=6+2=8\)
\(x=8:2=4\)
c) \(\left(2x-14\right)^7=128=2^7\)
\(2x-14=2\)
\(2x=14+2=16\)
\(x=16:2=8\)
d) \(x^4\cdot x^5=5^3\cdot5^6=5^4\cdot5^5\)
\(x=5\)
e) \(3\cdot\left(x+2\right):7\cdot4=120\)
\(x+2=120:3\cdot7:4\)
\(x+2=70\)
\(x=70-2=68\)
Lời giải:
a. $(x-1)^3=8=2^3$
$\Rightarrow x-1=2$
$\Rightarrow x=3$
b. $7^{2x-6}=49=7^2$
$\Rightarrow 2x-6=2$
$\Rightarrow 2x=8$
$\Rightarrow x=4$
c. $(2x-14)^7=128=2^7$
$\Rightarrow 2x-14=2$
$\Rightarrow 2x=16$
$\Rightarrow x=18$
d.
$x^4.x^5=5^3.5^6$
$x^9=5^9$
$\Rightarrow x=5$
e.
$3(x+2):7=120:4=30$
$3(x+2)=30.7=210$
$x+2=210:3=70$
$x=70-2=68$
1.Cho \(r\left(x\right)=-\left(3x-7\right)^2+2\left(3x-7\right)-17\)
Tìm GTLN của biểu thức r(x).
2. So sánh : \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)với \(B=3^{32}-1\)
3. Tìm x, y biết: \(y^2+2y+4x-2^{x+1}+2=0\)
Câu 3 kiểm tra lại đề lại với , nếu đúng thì phức tạp lắm, còn sửa lại đề thì là :
\(y^2+2y+4^x-2^{x+1}+2=0\)
\(=>\left(y^2+2y+1\right)+2^{2x}-2^x.2+1=0\)
\(=>\left(y+1\right)^2+\left(\left(2^x\right)^2-2^x.2.1+1^2\right)=0\)
\(=>\left(y+1\right)^2+\left(2^x-1\right)^2=0\)
Dấu = xảy ra khi :
\(\hept{\begin{cases}y+1=0\\2^x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=-1\\x=0\end{cases}}}\)
CHÚC BẠN HỌC TỐT...........
1, Khai triển ra ta được:
\(r\left(x\right)=-\left(9x^2-42x+49\right)+6x-14-17\)
\(=-9x^2+42x-49+6x-14-17\)
\(=-9x^2+48x-80\)
\(=-9x^2+48x-64-16\)
\(=-\left(\left(3x\right)^2-3x.2.8+8^2\right)-16\)
\(=-\left(3x+8\right)^2-16\)
\(Do-\left(3x+8\right)^2\le0\)
\(=>-\left(3x+8\right)^2-16\le-16\)
Dấu bằng xảy ra khi \(3x+8=0=>x=-\frac{8}{3}\)
Vậy giá trị nhỏ nhất là -16 tại \(x=-\frac{8}{3}\)
Tìm x thuộc N biết :
a) \(\left(\frac{3}{2}\right)^x=\frac{11.\left(-3\right)^{22}.3^7+9^3.3^{14}}{\left(4.3^{14}\right)^2}\)
b)\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{7}{9}\)
tìm x, y
\(a.\left(x-5\right)^2=\left(1-3x\right)^2\)
\(b.\left(x+5\right)^2+\left(3y-9\right)^4=0\)
\(c.\frac{1}{8}.16^x=2^x\)
\(d.2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
\(a,\left(x-5\right)^2=\left(1-3x\right)^2\)
\(\Rightarrow\left(x-5\right)^2-\left(1-3x\right)^2=0\)
\(\Rightarrow\left(x-5+1-3x\right)\left(x-5-1+3x\right)=0\)
\(\Rightarrow\left(-2x-4\right)\left(4x-6\right)=0\)
\(\Rightarrow\left(x+2\right)\left(2x-3\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+2=0\\2x-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=-2\\x=\frac{3}{2}\end{cases}}}\)
a) \(\left(x-5\right)^2=\left(1-3x\right)^2\)
\(\Rightarrow x-5=1-3x\)
\(\Leftrightarrow x+3x=1+5\)
\(\Leftrightarrow4x=6\)
\(\Leftrightarrow x=\frac{3}{2}\)