ai giải dùm với 35 – 3(x + 2) = 5
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\(TH1:x\ge0\)
\(\Rightarrow35-3x=5\left(2^3-4\right)\)
\(\Rightarrow35-3x=20\)
\(\Rightarrow3x=35-20\)
\(\Rightarrow3x=15\)
\(\Rightarrow x=5\left(tm\right)\)
\(TH2:x< 0\)
\(\Rightarrow35-3\left(-x\right)=5\left(2^3-4\right)\)
\(\Rightarrow35+3x=20\)
\(\Rightarrow3x=20-35\)
\(\Rightarrow3x=-15\)
\(\Rightarrow x=-5\left(tm\right)\)
Vậy \(S=\left\{5;-5\right\}\)
35 - 3|x| =5(8-4)
35-3|x|=5.4
35-3|x|=20
-3|x|=20-35
-3|x|=-15
|x|=-15: (-3)
|x|=5
\(\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
Vậy \(x\in\left\{5;-5\right\}\)
\(4x-x^2-5< 0\)
\(=\left(-x^2-4x\right)-5=-\left(x^2-2x.2+4-4\right)-5=-\left(x-2\right)^2+4-5\)
\(=-\left(x^2-2x\right)-1\)
Vì \(-\left(x^2-2x\right)\le0\)với mọi x nên \(-\left(x-2\right)^2-1< 0\)với mọi x
Vậy \(4x-x^2-5< 0\)với mọi x ( đpcm )
4x - x2 - 5 < 0 \(\forall\)x
Ta có : 4x - x2 - 5
= -x2 + 4x - 5
= - ( x2 - 4x + 5 )
= - ( x2 - 2.x.2 + 22 - 1 )
= - [( x - 2 )2 - 1 ]
Vì - ( x - 2 ) \(\le\)0 \(\forall\)x
\(\Leftrightarrow\)- ( x - 2 ) - 1 \(\le\)0 \(\forall\)x
Vậy .....
ĐKXĐ: \(\left\{{}\begin{matrix}2x+5>=0\\4-2x>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x>=-5\\2x< =4\end{matrix}\right.\Leftrightarrow-\dfrac{5}{2}< =x< =2\)
\(x^2+\sqrt{2x+5}+\sqrt{4-2x}=4x-1\)
=>\(x^2-4+\sqrt{2x+5}-3+\sqrt{4-2x}=4x-1-7\)
=>\(\left(x-2\right)\left(x+2\right)+\dfrac{2x+5-9}{\sqrt{2x+5}+3}+\sqrt{4-2x}=4x-8\)
=>\(\left(x-2\right)\left[\left(x+2\right)+\dfrac{2}{\sqrt{2x+5}+3}-4\right]+\sqrt{4-2x}=0\)
=>\(-\left(2-x\right)\left[\left(x-2\right)+\dfrac{2}{\sqrt{2x+5}+3}\right]+\sqrt{2\left(2-x\right)}=0\)
=>\(\sqrt{2-x}\left[-\sqrt{2-x}\left(x-2+\dfrac{2}{\sqrt{2x+5}+3}\right)+\sqrt{2}\right]=0\)
=>\(\sqrt{2-x}=0\)
=>x=2(nhận)
a)(6x2+13x-5):(2x+5)
=3x-1
b)(x3+8y3):(x+2y)
=x2(dư 8y3+2x2y)
tôi nghĩ là như thế
a) |x-3|=7
vì x<3=>x-3<0
=>|x-3|=3-x
=>3-x=7
=>x=4
b) x+|2-x|=6
vì x>2=>2-x<0
=>|2-x|=x-2
=>x+x-2=6
=>x=4
3 ( x - 2 ) - 150 = 24
( x - 2 ) - 150 = 24 : 3
( x - 2 ) - 150 = 8
x - 2 = 8 + 150
x - 2 = 158
x = 158 + 2
x = 160
Answer : 160
3/5 + x = 2/35 - 2/7
3/5 + x = -1/5
=> x = -1/5 - 3/5
=> x = -4/5
\(\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)
\(\frac{3}{5}+x=\frac{3}{35}-\frac{2}{7}\)
\(\frac{3}{5}+x=-\frac{1}{5}\)
\(x=-\frac{1}{5}-\frac{3}{5}\)
\(x=-\frac{4}{5}\)
#ủng_hộ_vs_ạ
#thanks_nhiều
35 – 3(x + 2) = 5
3(x + 2) = 35 - 5
3(x + 2) = 30
x + 2 = 30 : 3
x + 2 = 10
x = 10 - 2
x = 8
35-3(x+2)=5
3(x+2)=35-5
3(x+2)=30
x+2=30:3
x+2=10
x=10-2
x=8
Vậy x=8
HT