Tìm x biết (x2-17)*(x2-37)=<0
Tìm x, biết:
a) ( x + 3 ) 2 + (4 - x)(x + 4) = 1;
b) (2 - x) 3 +(3 +x)(9 - 3x + x 2 ) + 6x(1 - x) = 17;
c) x 4 - 2 x 2 +1 = 0.
a) Tìm được x = -4.
b) Tìm được x = 3.
c) Tìm được x = ±1.
Tìm x:
x2 + 5x = \(\sqrt{37}\)
\(x^2+5x=\sqrt{37}\)
\(\Leftrightarrow4x^2+20x=4\sqrt{37}\)
\(\Leftrightarrow4x^2+20x+25=4\sqrt{37}+25\)
\(\Leftrightarrow\left(2x+5\right)^2=4\sqrt{37}+25\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+5=\sqrt{4\sqrt{37}+25}\\2x+5=-\sqrt{4\sqrt{37}+25}\end{matrix}\right.\Leftrightarrow x=\dfrac{\pm\sqrt{4\sqrt{37}+25}-5}{2}\)
\(x^2+5x=\sqrt{37}\)
\(\Leftrightarrow x\left(x+5\right)=\sqrt{37}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{37}\\x+5=\sqrt{37}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{37}\\x=-5+\sqrt{37}\end{matrix}\right.\)
Vậy \(x=\sqrt{37};x=-5+\sqrt{37}\).
Tìm x, biết:
a ) x − 3 4 = 1 7 ; b ) − x + 3 5 + 13 20 = 5 6
c ) x 2 − 2 5 = 1 10 ; d ) 15 x − 1 3 = 28 51
a ) x = 25 28 b ) x = 5 12 .
c) x= 1 d) x = 17
a; \(x\) - \(\dfrac{3}{4}\) = \(\dfrac{1}{7}\)
\(x\) = \(\dfrac{1}{7}\) + \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{4}{28}\) + \(\dfrac{21}{28}\)
\(x\) = \(\dfrac{25}{28}\)
Vậy \(x=\dfrac{25}{28}\)
b; - \(x\) + \(\dfrac{3}{5}\) + \(\dfrac{13}{20}\) = \(\dfrac{5}{6}\)
(\(\dfrac{12}{20}\) + \(\dfrac{13}{20}\)) - \(x\) = \(\dfrac{5}{6}\)
\(\dfrac{5}{4}\) - \(x\) = \(\dfrac{5}{6}\)
\(x\) = \(\dfrac{5}{4}\) - \(\dfrac{5}{6}\)
\(x\) = \(\dfrac{30}{24}\) - \(\dfrac{20}{24}\)
\(x\) = \(\dfrac{5}{12}\)
Vậy \(x=\dfrac{5}{4}\)
a x2 +2x√x+1=8x-1
b x+√9-x2 -x√9-x2=3
c x2+x+12√x+1=36
d x+√17-x2 + x√17-x2=9
Cho x 1 là giá trị thỏa mãn 3 7 + 1 7 : x = 3 14 và x 2 là giá trị thỏa mãn 5 7 + 2 7 : x = 1 . Khi đó, chọn câu đúng
A. x 1 = x 2
B. x 1 < x 2
C. x 1 > x 2
D. x 1 = 2 x 2
Tìm X
3 x + 6 = 37
25 x2 = 25
8x - 3 = 1
\(25\cdot x^2=25\\ x^2=25\div25\\ x^2=1\\ x^2=1^2\\ x=1\\ 8^{x-3}=1\\ 8^{x-3}=8^0\\ x-3=0\\ x=0+3\\ x=3\)
`@` `\text {Ans}`
`\downarrow`
\(3^{x+6}=37\)
`=>`\(3^x\cdot3^6=37\)
`=>`\(3^x=37\div3^6\)
`=>` \(3^x=\dfrac{37}{729}\)
Bạn xem lại đề.
\(25x^2=25\)
`=>`\(x^2=25\div25\)
`=>`\(x^2=1\)
`=> x=1`
\(8^{x-3}=1\)
`=>`\(8^x\div8^3=1\)
`=>`\(8^x=8^3\)
`=> x=3`
Tìm x, biết:
a) ( x - 4 ) 2 - (x - 2)(x + 2) = 6;
b) ( x - 1 ) 3 + (2 - x)(4 + 2x + x 2 ) + 3x(x + 2) = 17.
a) Tìm được x = 7 4 . b) Tìm được x= 10 9 .
Tìm x biết:
a) [( x+ 32) -17] x2= 42
b) 125+(145-x)= 175
3X x(x-5)= 0
a) \(\left[\left(x+32\right)-17\right].2=42\)
\(\left[\left(x+32\right)-17\right]=42:2\)
\(\left[\left(x+32\right)-17\right]=21\)
\(\left(x+32\right)=21+17\)
\(\left(x+32\right)=38\)
\(x=38-32\)
\(x=6\)
b) \(125+\left(145-x\right)=175\)
\(\left(145-x\right)=175-125\)
\(\left(145-x\right)=50\)
\(x=145-50\)
\(x=95\)
Tìm m để phương trình x2 + (4m + 1)x + 2( m - 4 ) = 0 có hai nghiệm x1 , x1 thỏa mãn | x1 - x2 | = 17
\(\text{Δ}=\left(4m+1\right)^2-8\left(m-4\right)\)
\(=16m^2+8m+1-8m+32\)
\(=16m^2+33>0\)
Do đó: Phương trình luôn có hai nghiệm phân biệt
Ta có: \(\left|x_1-x_2\right|=17\)
\(\Leftrightarrow\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=17\)
\(\Leftrightarrow\sqrt{\left(4m+1\right)^2-4\cdot2\cdot\left(m-4\right)}=17\)
\(\Leftrightarrow\sqrt{16m^2+8m+1-8m+32}=17\)
\(\Leftrightarrow16m^2+33=289\)
=>m=4 hoặc m=-4
Bài 6. Tìm x Z, sao cho:
a. (x – 3) – 12 = – 25
f. 121 – (35 – x) = 2.52
c. 72 – (84 – 9x) : 7 = 69
g. 2(17 + x) – (400 – 325) = – 31
d. (x + 7).(x – 5) =0
h. (x – 10).(x2 – 9) = 0
e. (x + 8).(x2 + 1) = 0
i. 17 – { – x + [– x – (– x)]} = – 16
a: \(\Leftrightarrow x-3=-13\)
hay x=-10